
PRESENTED BY 



894 



(p. £.Q^te^. 



ELEMENTS 



DIFFERENTIAL AND INTEGRAL 
CALCULUS, 



EXAMPLES AND APPLICATIONS. 



BY 

JAMES M: TAYLOR, A.M., 

PROFESSOR OF MATHEMATICS, COLGATE UNIVERSITY. 



>^/ 



BOSTON, U.S.A.: 

PUBLISHED BY GINN & COMPANY. 

1894. 




Entered according to, Act of Congress, in the year 1884, by 

JAMES M. TAYLOR, 
in the Office of the Librarian of Congress, at Washington. 



Entered according to Act of Congress, in the year 1891, by 

JAMES M. TAYLOR, 
in the Office of the Librarian of Congress, at "Washington. 



Sift * ' 
Edwn L. Whitney 

DEC 8- 1938 



Typography by J. S. Cushing & Co., Boston. 



Presswork by G-inn & Co., Boston. 



PEEFACE 



THE object of the following treatise is to present sim- 
ply and concisely the fundamental problems of the 
Calculus, their solution, and more common applications. 

Since variables are its characteristic quantities, the 
first fundamental problem of the Calculus is, To find the 
ratio of the rates of change of related variables. To ena- 
ble the learner most clearly to comprehend cnis problem, 
the author has employed the conception of rates, which 
affords finite differentials and the simplest demonstration 
of many principles. The problem of Differentiation hav- 
ing been clearly presented, a general method of its solu- 
tion is obtained by the use of limits. This order of 
development avoids the use of the indeterminate form -, 
and secures all the advantages of the differential nota- 
tion. Many principles are proved, both by the method 
of rates and that of limits, and thus each is made to 
throw light upon the other. 

In a final chapter, the method of infinitesimals is briefly 
presented; its underlying principles having been previ- 
ously established. 

The chapter on Differentiation is followed by one on 
Integration ; and in each, as throughout the work, there 



IV PREFACE. 

are numerous practical problems in Geometry and Me- 
chanics, which serve to exhibit the power and use of 
the science, and to excite and keep alive the interest 
of the student. 

In writing this treatise, the works of the best Ameri- 
can, English, and French authors have been consulted ; 
and from these sources the most of the examples and 
problems have been obtained. 

The author is indebted to Professors J. E. Oliver and 
J. McMahon of Cornell University, and Professor O. 
Root, Jr., of Hamilton College, for valuable suggestions; 
and to Messrs. J. S. Gushing & Co. for the typograph- 
ical excellence of the book. 

J. M. TAYLOR. 
Hamilton, N.Y., 

Nov.. 1884. 



CONTENTS. 



CHAPTER I. 

INTRODUCTION. 
Section. Page. 

1. Definition of variable and constant 1 

2. Definition of function and independent variable 1 

3. Classification of functions 2 

4. Definition of continuous variable and continuous function 2 

5. Definition of the limit of a variable 3 

6. Limits of equal variables 3 

7- Limit of the product of a constant and a variable ...,.,. 4 

8. Limit of the product of two or more variables ........ 4 

9. Limit of the quotient of two variables 4 

10. Limit of the sum of two or more variables 4 

11. Definition of uniform change 5 

12. Definition of increment . 5 

13. Definition of differential 5 

14. Illustrations of differentials 6 

15. Definition of inclination, slope, and tangent 7 

16. Geometric signification of -^ 7 

dx 

17. Limit of the ratio of the increments of y and x 8 



CHAPTER II. 

DIFFERENTIATION. 

18. Definition of differentiation. Differentiation of ax 2 10 

Algebraic Functions. 

19. Differential of the product of a constant and variable 10 

20. Differential of a constant 11 

21. Differential of the sum of two or more variables ...... 11 



VI CONTENTS. 

Section- Page. 

22. Differential of the product of two variables 12 

23. Differential of the product of several variables 13 

24. Differential of a fraction 14 

25. Differential of a variable with a constant exponent 14 

26. General symbol for the differential off(x). Examples .... 15 

27. Definition of an increasing and a decreasing function 17 

28. Definition of derivative 17 

29. Measure of rate of change 18 

30. Signification of ^ 18 

dt 

31. Signification off'(x) or -1. . 18 

dx 

32. Limit of the ratio of Ay to Ax. Applications 19 

33. Definition of velocity and acceleration. Examples 23 

Logarithmic and Exponential Functions. 

34. Differential of a logarithmic Junction 24 

35. The greater the base, the smaller the modulus • . 25 

36. Naperian system 25 

37. Differential of c* . . 2G 

38. Differential of if . 26 

39. Logarithmic differentiation. Examples . 26 

Trigonometric Functions. 

40. Definition of the unit of angular measure 29 

41. Differential of sin x and cos x 29 

42. Differential of tan x . . 30 

43. Differential of cot x 30 

44. Differential of sec x 30 

45. Differential of cosec x 31 

46. Differential of vers x 31 

47. Differential of covers x 31 

48. Limit of the ratio of an arc to its chord 31 

49. Differentiation of sin a: by the method of limits. Examples . . 32 

Anti-Trigonometric Functions. 

50. Differential of sin -1 * 35 

51. Differential of cos -1 * 35 

52. Differential of tan -1 * 35 

53. Differential of cot -1 * 36 

54. Differential of sec -1 * . 36 



CONTENTS, Vll 

Section. Page. 

55. Differential of cosec" 1 ^ ■» . 36 

56. Differential of vers _1 x 36 

57. Differential of covers" 1 ^. Examples 36 

Miscellaneous examples 39 



CHAPTER in. 

INTEGRATION. 

58. Definition of integral and integration. Sign of integration ... 43 

59. Elementary principles 43 

60. Fundamental formulas 44 

61. Statement of formulas 1 and 2. Examples 46 

62. Auxiliary formulas. Examples 50 

63. Trigonometric differentials. Examples 55 

64. Definite integrals. Examples 58 

Applications to Geometry and Mechanics. 

65. Rectification of curves. Examples 60 

6Q. Areas of plane curves. Examples . 61 

67. Graphical representation of any integral . , 63 

68. Areas of surfaces of revolution. Examples 63 

69. Volumes of solids of revolution. Examples 65 

70. Fundamental formulas of mechanics. Examples 6Q 

1. Formulas for uniformly accelerated motion. 

2. Motion down an inclined plane. 

3. Motion down a chord of a vertical circle. 

4. Values of v and s when a varies directly as t. 

5. Geometrical representation of the time, velocity, distance, 

and acceleration. 

6. Path of a projectile. 

7. Path, velocity, and acceleration of a hody whose velocity 

in each of two directions is given. 

CHAPTER IV. 

SUCCESSIVE DIFFERENTIATION. 

71. Successive derivatives „ 71 

72. Signification oif>'(x),f'"{x),f»(x). Examples 71 

73. Successive differentials 73 

74. Relations between successive differentials and derivatives. 

Examples 73 



Vlil CONTENTS. 



CHAPTER V. 

SUCCESSIVE INTEGRATION AND APPLICATIONS. 

Section. Page. 

75. Successive integration. Examples 76 

76. Problems in mechanics 77 

1. Time and velocity of a falling body below the surface of 

the earth. 

2. Maximum velocity with which a falling body can reach the 

earth. 

3. Velocity of a body falling from the sun. 

4. Velocity of a body falling in the air. 

5. Velocity of a body projected into a medium. 

6. Velocity of a body sliding down a curve. 

7. The cycloidal pendulum isochronal. 

8. The length and equation of the catenary. 



CHAPTER VI. 

INDETERMINATE FORMS. 

77. Value of functions assuming an indeterminate form. Examples 86 

Evaluation by Differentiation. 

78. The form -. Examples 87 

79. The form -g. Examples 88 

80. The forms • go and co — go. Examples 89 

81. Eunctions whose logarithms assume the form ± • go. Examples 91 

82. Compound indeterminate forms. Examples 91 

83- Evaluation of derivatives of implicit functions. Examples . . 92 

CHAPTER VII. 

DEVELOPMENT OF FUNCTIONS IN SERIES. 



94 



84. Definition of series, convergent infinite series, and sum of infinite 

series 

85. Definition of the development of a function 94 

86. Definition of Taylor's formula 95 

87. Proof of Taylor's formula 95 

88. Proof of Maclaurin's formula 96 



CONTENTS. ix 

Section. Page. 

89. Proof of the binomial theorem . 97 

90. Development of log a (:r + y) . . 97 

91. Development of a x +'J 98 

92. Development of (a -f x) m 98 

93. Development of sin x . . . „ . 98 

94. Development of cos x 98 

95. Exponential series 99 

96. Logarithmic series 99 

97. Failure of Taylor's and Maclaurin's formulas 102 

98. Lemma '. 102 

99. Completion of Taylor's and Maclaurin's formulas 103 

100. A second complete form 104 

101. Proof that $2 = as n = oo 105 

\n 

102. Maclaurin's formula develops a x 105 

103. Maclaurin's formula develops sin x and cos x 106 

104. The logarithmic series holds for x>—l and <+l 106 

105. Development of (I + x) m holds for x>—l and <+l 107 

106. One form of the development of (a + x) m always holds .... 108 

107. Development of tan -1 x, and value of it 109 

108. Development of sin -1 ^, and value of it 109 

109. Geometric proof that /(a + h) =J\a) + hf'{a + Bit). Examples 110 



CHAPTER VIII. 

MAXIMA AND MINIMA. 

110. Definition of maximum and minimvm 112 

111. The critical values of a; 112 

112. Two methods of examining^ x ) at critical values of x . . . . 113 

113. Maxima and minima occur alternately 114 

114. Principles facilitating the solution of problems ....... 114 

Examples and geometric problems 115 

CHAPTER IX. 

FUNCTIONS OF TWO OR MORE VARIABLES, AND CHANGE OF THE 
INDEPENDENT VARIABLE. 

115. Applicability of previous rules for differentiation 125 

116. Definition of partial differential 125 

117. Definition of total differential 125 



X CONTENTS. 

Section. Page, 

118. Definition of partial derivative 125 

119. Definition of total derivative 125 

120. The value of the total differential 126 

121. The signification of partial derivatives. Examples ...... 126 

122. One method of finding the total derivative 127 

123. Formulas for finding the total derivative. Examples .... 128 

124. Implicit functions 130 

125. Formula for the derivative of an implicit function. Examples 130 

126. Successive derivatives of an implicit function. Examples . . 131 

127. Successive partial differentials and derivatives ....... 132 

128. The order of differentiations is indifferent. Examples .... 133 

129. Formulas for successive differentials 134 

130. Change of the independent variable 135 

131. Forms of successive derivatives, dx being variable. Examples 135 

CHAPTER X. 

TANGENTS, NORMALS, AND ASYMPTOTES. 

132. The equation of a tangent 139 

133. The equation of a normal. Examples 139 

134. Length of tangent, normal, subtangent, and subnormal. Exam- 

ples .... 141 

135. Important principle in the method of limits 143 

136. Length of polar subtangent, subnormal, etc 143 

137. A second proof. Examples 145 

138. Rectilinear asymptotes. Examples 146 

139. Asymptotes determined by inspection or expansion. Examples 149 

140. Asymptotes to polar curves. Examples 151 



CHAPTER XL 

DIRECTION OF CURVATURE, SINGULAR POINTS, AND CURVE 
TRACING. 

141. Direction of curvature. Examples 154 

142. Definition of singular points 156 

143. Points of inflexion. Examples 156 

144. Points of inflexion on polar curves 157 

145. Definition of multiple point 157 

146. At a multiple point -JL has two or more values 158 

dx 



CONTENTS. XI 

Section. , ^ Page. 

147. At a multiple point -1. assumes the form - 158 

dx 

148. Examination of a curve for multiple points. Examples . . . 159 

149. Shooting points and stop points 163 

150. Curve tracing. Examples . . . • • 163 

151. Tracing polar curves. Examples 168 



CHAPTER XII. 

CURVATURE, EVOLUTES, ENVELOPES, AND ORDER OF CONTACT. 

152. Direction of a curve 170 

153. Definition of curvature 170 

154. Measure of the curvature of a circle 170 

155. Eormula for curvature 171 

156. Eadius of curvature 171 

157. Eadius of curvature in polar curves 172 

158. Intersection of a curve and its circle of curvature ....... 172 

159. Exception to § 158 172 

160. Definition of an e volute ~ . 174 

161. Deduction of the equation of an e volute. Examples 174 

162. A normal to an involute is a tangent to its evolute 177 

163. Eength of an arc of an evolute 178 

164. Tracing an involute from its evolute 179 

165. Definition of a variable parameter 180 

166. Definition of an envelope 180 

167. Deduction of the equation of an envelope . 180 

168. An envelope is tangent to each curve of the series. Examples 181 

169. Definition of contact of different orders 183 

170. Intersection of curves at their point of contact 183 

171. Osculating curves , 184 

Examples 185 



CHAPTER XIII. 

INTEGRATION OF RATIONAL FRACTIONS. 

172. Decomposition of rational fractions c . . . . 188 

173. Simple factors of the denominator real and unequal. Exam- 

ples . 188 

174. Some of the simple factors real and equal. Examples .... 190 

175. Some imaginary and unequal Examples 191 

176. Some imaginary and equal. Examples , . , . . 193 



Xll CONTENTS. 

CHAPTER XIV. 

INTEGRATION BY RATIONALIZATION. 

Section. Page. 

177. Rationalization of a differential „ 195 

c 

178. Differentials containing surds of the form xl. Examples . . 195 

c 

179. Surds of the form (a+ bx)a. Examples 196 

180. Surds of the form Va + bx + x 2 196 

181. Surds of the form Va + bx - x 2 197 

182. Binomial differentials 199 

183. Conditions of rationalization of x m (a + bx n )ldx. Examples . 199 

CHAPTER XV. 

INTEGRATION BY' PARTS AND BY SERIES. 

184. Formula for integration by parts. Examples 203 

185. Formulas of reduction. Examples 204 

186. Integration of (j>(x)(logx) n dx. Examples . 210 

187. Integration of x n a mx dx. Examples 211 

188. Integration of sin TO :r cos w :r dx. Examples 211 

189. Integration of x n sin ax dx and x n cos ax dx. Examples . . . 213 

190. Integration of e ax &m. n xdx and e ax cos n xdx. Examples. . . . 213 

191. Integration of — 215 

a -f b cos x 

192. Integration of f{ x) sin^xdx, etc. Examples ........ 215 

193. Integration by series. Examples 216 

194. Integration often leads to higher functions 217 

CHAPTER XVI. 

LENGTH AND AREAS OF PLANE CURVES, AREAS OF SURFACES OF 
REVOLUTION, VOLUMES OF SOLIDS. 

195. Examples in rectification of plane curves 218 

196. Rectification of polar curves. Examples . .- . 219 

197. Examples in quadrature of plane curves „ . . 220 

198. Quadrature of polar curves. Examples 222 

199. Examples in quadrature of surfaces of revolution 223 

200. Examples in cubature of solids of revolution 224 

201. Equations of curves deduced by aid of the Calculus. Exam- 

ples ...... 225 



CONTENTS. Xlll 



CHAPTER XVn. 

THE METHOD OF INFINITESIMALS. 

Section, Page. 

202. Definition of infinitesimals and infinites 226 

203. Orders of products and quotients 227 

204. Geometric illustration of infinitesimals of different orders . . 227 

205. First fundamental principle of infinitesimals 228 

206. Eule for differentiation 229 

207. Second fundamental principle of infinitesimals 231 

208. Integration as a summation 232 

209. Definition of centre of gravity 233 

210. Centre Of gravity of any plane surface 233 

211. Centre of gravity of any plane curve 234 

212. Centre of gravity of any solid of revolution 234 

Examples , r . . . . 235 



ELEMENTS OP THE CALCULUS 



Elements of the Calculus. 



CHAPTER I. 



INTRODUCTION. 



1. In the Calculus there are two kinds of quantities considered, 
variables and constants. 

A Variable is a quantity that is, or is conceived to be, con- 
tinually changing in value. Variables are usually represented 
by the final letters of the alphabet. 

A Constant is a quantity whose value is fixed or invariable. 
Constants are usually represented by figures or the first letters 
of the alphabet. Particular values of variables are constants. 

In the Calculus the locus of an equation is conceived as traced 
by a moving point called the Generatrix. If a = ob, the locus 
of a^-f y 2 =a 2 is the circle abcd. Now, 
as the generatrix traces this circle, its 
coordinates, x and y, continually change 
in value, and are therefore variables ; 
while a retains the value ob, and is 
therefore a constant. 

2. Functions and Independent Vari- 
ables. One variable is & function of an- 
other, when the two are so related that 
any change of value in the second produces a change of value 
in the first. 

For example, the area of a varying square is a function of 
its side ; the volume of a variable sphere is a function of its 
radius ; all mathematical expressions depending on x for their 
values, as ax 3 , bx^-j-cx 2 , sina, log.T, etc., are functions of a?. 

An independent variable is one to which any arbitrary value 
or law of change may be assigned ; as, x in x?, x in sin$, etc. 




Z INTRODUCTION. 

The symbol f(x) is used to denote any function of x, and is 
read " function of a?." When several functions of x occur in 
the same investigation, we employ other symbols, as f'(x), 
F(x), <£(#), etc., which are read u f prime function of x" 
" F function of x" " </> function of x," etc. According to this 
notation, y =f(x) represents any equation between x and y, 
when solved for y. 

3. Algebraic and Transcendental Functions. — An algebraic 
function is one that is expressed in terms of its variable or 
variables, by means of algebraic signs, without the use of 
variable exponents; as, ax s — 2 ex 2 , ox 2 — x, etc. 

All functions not algebraic are called transcendental. These 
are sub- divided into exponential, logarithmic, trigonometric, and 
anti-trigonometric . 

An Exponential function is one in which the variable enters 
the exponent ; as, a cx , y ax . 

A Logarithmic function is one that involves the logarithm of 
a variable ; as, log x, log (bx + c). 

The sine, cosine, tangent, etc., of a variable angle are called 
Trigonometric functions. 

The symbol sin -1 a?, read "anti-sine of x," denotes the angle 
whose sine is x. Sin -1 #, cos"" 1 ^, tan -1 ^, etc., are called Inverse 
Trigonometric, or Anti-Trigonometric, functions. 

4. A variable is Continuous, or varies continuously, when, in 
passing from one value to another, it passes successively through 
all intermediate values. 

A Continuous function is one that is constantly real, and 
varies continuously, when its variable varies continuously. 
Some functions are continuous for all real values of their vari- 
ables, others only for those between certain limits. Thus, if 
y = ax -f- b, or y = sin x, y is evidently a continuous function of 
x for all real values of x ; but, if y = ± Vr 2 — x 2 , y is continuous 
only for values of x between the limits — r and -f- r. 

The Calculus treats of variables and functions only between 
their limits of continuity ; hence all the values of x and f(x) 
that it considers are represented geometrically by the coordi- 
nates of the points of the plane curve whose equation is yz=f(x). 



'THEORY OF LIMITS. 3 

Theory of Limits. 

5. For convenience of reference, we give here a brief state- 
ment of the theory of limits. 

The Limit* of a variable is a constant quantity which the 
variable, in accordance with its law of change, approaches 
indefinitely near, but which it never reaches. The variable may 
be less or greater than its limit. 

Thus, if the number of sides of a regular polygon inscribed 
in or circumscribed about a circle be indefinitely increased, 
the area of the circle will be the limit of the area of either 
polygon, and the circumference will be the limit of the peri- 
meter of either. When the polygons are inscribed, the 6 variable 
area and perimeter are less than their limits ; and, when the poly- 
gons are circumscribed, the variable area and perimeter are 
greater than their limits. 

By increasing the number of terms, the sum of the series, 
1 -j- -g- + i -f- -J -f- etc. , can be made to approach 2 as nearly as we 
please, but it cannot reach 2 ; hence 2 is the limit of the sum. 

Again, if a point starting from a move the distance ac 
(=-Jab) the first second, the distance 
cd (=A-cb) the second second, and so ' ' ' ' 

^ z ' A C D B 

on, ab will evidently be the limit of the Fi 2> 

line traced by this point. 

Cor. The difference between a variable and its limit is a vari- 
able whose limit is zero. 

6. If two variables are continually equal, and each approaches 
a limit, their limits are equal; that is, ?/x = y, and limit (x) 
= a, and limit (y) = b, a = b. 

For, since x=y, a — x — a — y\ hence, as a is the limit of x, it 
is also of y (§ 5, Cor.). Since a and b each is a limit of 'y, and 
y cannot approach two unequal limits at the same time, a = b. 

Cor. If one of two continually equal variables approaches a 
limit, the other approaches the same limit. 

* The student should carefully note the two senses in which the word 
limit is used. In the theory of limits, a limit is a value which the variable 
cannot reach ; in other cases, as in § 4, a limit is the greatest or the least 
value which the variable actually reaches. 



4 INTRODUCTION. 

7. The limit of the product of a constant and a variable is the 
product of the constant and the limit of the variable; that is, if 
limit (x) = a, limit (ex) = ca. 

Let v = a —: x ; 

then cx= ca — cv. 

Now limit (cv) = 0, since limit (v) = ; 

hence limit (ex) = limit (ca — cv) = ca. 

8. The limit of the variable product of two or more variables 
is the product of their limits; that is, if limit (x) = a, and limit 
(y) = b, limit (xy) = ab. 

Let v== a~x, and v x = b — y ; 

then , x = a — v, and y — b — ^ ; 
.*. xy = ab — (av ± -\- bv — W]) . 

Now limit (av x -\- bv - 7 -vv i )*=0; 
hence limit (xy) = limit \_ab — (a^-f- bv — vv 1 )~\ = ab. 

In like manner, the theorem is proved for n variables. 

9. The limit of the variable quotient of two variables is the 
quotient of their limits; that is, if limit (x) = a, and limit (y) =b, 
limit (x -f- y) = (a -r- b) .f 

Let z — x-^y, 

and c = limit (z) , or limit (x-v-y). 

Then a? = y« ; .-. a = bc ; §§ G, 8. 

.*. limit (x -s-2/)[= c] = a-f- 5. 

10. T%e /imrt q/*£7ie variable sum of a finite number of vari- 
ables is the sum of their limits ; that is, if limit (x) = a, limit 
(y) = b, ZiwiiY (z) = c, etc., 

limit (x+y + z-| ) = a + b + cH 

Let v = a — x, v 1 = b — y, v 2 = c — z, etc. 
Then x -f y -f z -\ 

= (a + & + c + .»)-(« + v 1 + v, + --); 

.-. limit (a) + ?/ + 2 + •••) 

= limit[(a + & + c + ...)-(v + ^ + v 2 +...)] 
= a + 6 -fc H 

* When u and v x have unlike signs, the difference, av x -\- bv — vv v may be- 
come zero for particular values of v and v v but it cannot remain zero, since 
xy is variable. The same is true of the difference, v + i\ + v 2 + • ••, in § 10. 

t This principle does not hold when the limit of the divisor is zero. 



INCREMENTS AND DIFFERENTIALS. O 

Cor. Wlien the product, quotient, or sum of two or more vari- 
ables is equal to a constant, the product, quotient, or sum of their 
limits is equal to the same constant. 

11. The Change of a variable is Uniform, when its value 
changes equal amounts in equal arbitrary portions of time. In 
all other cases the change is variable. 



Thus, if from a toward b a point move ^ — ^ — J — c — ^ — e — ^ 
equal distances, as aci, ab, be, etc., in equal Fig 3 

arbitrary portions of time, the increase of 

the line traced will be uniform. Again, if the motion of a point 
along a straight line be uniform, the change of each of its 
rectilinear coordinates will evidently be uniform. 

12. An Increment of a function or variable is the amount of 
its increase or decrease in any interval of time, and is found by 
subtracting its value at the beginning of the interval from its 
value at the end. Hence, if a variable is increasing, its incre- 
ment is positive ; and, if it is decreasing, its increment is 
negative. An increment of a variable is denoted by writing 
the letter A before it; thus, Ax, read "increment of x" is the 
symbol for an increment of x. If y =f(x) , Ax and Ay repre- 
sent corresponding increments, that is, the increments of x and 
y in the same interval of time. 

Let oph be the locus of y =f(x) referred to the rectangular 
axes ox and or. If, when x = oa, 
Ax = ob — oa = ab ; then 
Ay =bp f — ap = ep' ; if, when x = oc. 
Ax = cf ; then Ay = fh — cd = — xd. 

In the last case Ay is negative, but o a b c f 
it is properly called an increment, Fig - 4 - 

since it is what must be added to the first value to produce 
the second. 

13. The Differential of a function or variable at any value is 
what ivould be its increment in any interval of time, if at that 
value its change became uniform. Hence, the differential of a 



{ . 

p 


E 


X 




H 


p 






\ 






V 



6 



INTRODUCTION. 



variable is positive or negative, according as the variable is 
increasing or decreasing. The interval of time, though arbitrary, 
must be the same for a function as for its variable. 

If the change of a variable be uniform, any actual increment 
may evidently be taken as its differential. 

The differential of a variable is represented by writing the 
letter d before it ; thus, dx, read " differential x," is the symbol 
for the differential of x. When the symbol of a function is not 
a single letter, parentheses are used; thus, d(x?) and d(x? — 2x) 
denote the differentials of x 3 and x 2 — 2x. 




14. Illustrations of Differentials. Conceive a variable right 

triangle as generated by the perpendicular moving uniformly to 

the right. Let y represent its area, x its 

base, and 2 ax its altitude; then y=a>x 2 . Let 

bh be Ax estimated from the value ab ( = x') , 

then bhmc will be Ay. But, if the increase 

of the area became uniform at the value abc, 

the increment of the area in the same time 

would evidently be bhoc ; hence, bhoc and 

bh may be taken as the differentials of y and 

x, when x— x'. But bhoc = 2ax'dx, hence, 

in general, dy[_ = d(ax 2 )~]=2axdx. lfa=l,y=x 2 , &nddy=2xdx. 

Here Ay = dy-\- triangle com. 

The signification of dy=2ax'dx is evidently that, when 
x = x\ y, the area, is changing in units of surface 2 ax' times as 
fast as x is in linear units. 

Again, let opn be the locus of y=f(x), referred to the axes 
ox and oy. Conceive the area between ox and 
the curve as traced by the ordinate of the curve 
moving uniformly to the right. Let z repre- 
sent this area, and let ab be Ax estimated 
from the value oa (=#'); then abp'p = Az. 
But, if the increase of z became uniform at the 
value oap, its increment in the same interval 
would evidently be abdp ; hence ab and abdp may be taken as 
the differentials of x and z respectively, when x = x f . 




GEOMETRIC ILLUSTRATIONS OF DIFFERENTIALS. i 

Hence dz = abdp =afcIx = y'dx ; or, in general, dz = ydx, 
which evidently means that z is changing y times as fast as x. 

Area above the axis of x being positive, area below it is 
negative ; hence, where the curve lies below the axis of x, the 
area decreases as x increases, and ydx is negative as it should be. 

Here Az = dz -\- area pdp\ 

15. The Inclination of a straight line referred to rectangular 
axes is the angle included between the axis of abscissas and the 
line. The direction of a line with respect to the axis of x is 
determined b}~ its inclination. 

The Slope of a line is the tangent of its inclination. Thus, in 
Fig. 7, hzp is the inclination of za, and tanHzp is the slope of za. 

The direction of motion of the generatrix of a straight line is 
constant? while the direction of motion of the generatrix of a 
curve is variable. 

A Tangent to a curve at any point is the straight line that 
passes through that point, and has the same direction as the 
curve at that point ; or, a tangent to a curve at any point is the 
straight line that the generatrix would trace, if its direction of 
motion became constant at that point. The slope of a curve at 
any point is the slope of its tangent at that 
point. Thus, if, in Fig. 7, pa is a tangent to 
the curve at p, tanHZA is the slope of the 
curve at p. 

16. Geometric Signification of JL. Let mn 

dx 

be the locus of y=f{x), and let x' be the 
abscissa of any point upon it, as p. If at p JTz j- h x . 
the motion of the generatrix of the curve Fi §- <• 

became uniform along the tangent pa, it is 
evident Jthat the change of each of its coordinates would also 
become uniform. Hence pe, ea, and pa may be taken respec- 
tively as the differentials of x, y, and the length of the curve, 
When x = x' ; for they are what would be the simultaneous 
increments of these variables, if the change of each became 

dll EA 

Uniform at the value considered. Therefore -f- = — — tan epa 

dx PE 



INTRODUCTION. 



= tan hz a, which is the slope of the curve at p. Hence, in 

general, -JL is the slope of the curve y = f (x) at any point (x, y). 
dx 

Cor. 1. If ea, or dy, be c times as great as pe, or dx, y is 
evidently increasing c times as fast as x, when x = 01. 

Cor. 2. If s represent the length of the curve mn, PA = ds, 
and ds 2 = dx 2 + dy 2 , in which ds 2 denotes the square of ds. 

17. Limit of the Ratio of the Increments of y and x. 
Let mn be the locus of y =f(x) , and ed, a tangent at p, any 

point upon it ; then the slope of 




this tangent 



mn [= pc] = Ax, when estimated 

from the value om, then cp'=A?/. 

Draw the secant pp' ; then 

-^- = the slope of the secant pp'. 

Ax 

Conceive Ax to approach as 
its limit ; then the slope of the 
secant will approach the slope of the tangent as its limit.* 



limit 
A.r = 



'Ay 

Ax 



■\ = dy t 
dx 



Hence, the ratio of the differential of a function to that of its 
variable is the limit of the ratio of their increments, as these incre- 
ments approach zero as their limit. 



dy 



limit 



Ay 

Ax 



is finite, except 



Cor. 1. It is evident that — , or 

dx Ax = ° 

where the locus of y=f(x) is parallel or perpendicular to the 
axis of x, where it is or go. 



* This statement, if not sufficiently evident, may be demonstrated as 
follows : When the arc pap' is continuous in curvature, and this arc can 
always be made so small that it will be continuous, the slope of the secant 
pp' is equal to the slope of a tangent to the arc pap' at some point, as a. 
Now, as the arc pap' approaches zero as its limit, the point a approaches p 
as its limiting position ; hence the slope of the secant pp' approaches the 
slope of the tangent pd as its limit. 



x limit 
I A* = 



read " the limit of — as Ax approaches as its limit. 
Ax 



LIMIT OF THE RATIO OF INCREMENTS. 



Aw 

Cor. 2. If — ^ be constant, the locus of y=f(x) is evidently 

a straight line, in which case — = -^. 

° Ax dx 

Cor. 3. A tangent to ran at p is evidently the limiting posi- 
tion of the secant pp' as p' approaches p and arc p'p = 0. 

The following is another proof of the important principle 
established above : — 

Second Proof.* Conceive the area between ox and the 
curve opft (Fig. 9) as traced by the ordinate 
of the curve moving to the right. Let z T 
represent this area, and let ab be Ax esti- 
mated from the value oa ; then Ay = dp', and 

Az = ABP'P. 

NOW ABDP < ABP'P < ABP'M ; 

.-. y'Ax <Az< (y'+ Ay) Ax. 
Dividing by Ax, we have 

, Az , 

Az 
Whence — differs from y' less than y'-\- Ay does ; but 





M 


p v^ 


p 


/S\ 







A B 

Fig. 9. 



Ax 

limit 
Ax 



[7/+A2/] = 2/'; 



limit 
Ax = 



Az 
Ax 



= y 



But 
Hence 



dz 
dx 



= y 



§ 14 



limit 



Az 
Ax 



dz 
dx 



* This demonstration assumes that any function of x may be repre- 
sented graphically by the area between a curve and the axis of x. That 
many functions of x may be thus represented is very evident, and that any 
may be follows from § 67. 



CHAPTER II. 
DIFFERENTIATION. 

18. Differentiation is the operation of finding the differential 
of a function. The sign of differentiation is d ; thus d in d(x 3 ) 
indicates the operation of differentiating x\ while the whole 
expression d(x") denotes the differential of x" (see § 13). 

To differentiate ax 2 , let y = ax 2 , and let x' and y' be any cor- 
responding values of x and y ; then 

y'= ax' 2 . (1) 

Let Ax be any increment of x estimated from the value x\ 
and Ay the corresponding increment of y ; then 

y'+Ay = a(x'+Ax) 2 = ax'' 2 + 2 ax' Ax + a{Ax) 2 . (2) 
Subtracting (1) from (2), we have 

Ay = 2 ax' Ax -f a (Ax) 2 , or ^ = 2 ax' + aAx. (3) 

Ljk-X 



. limit 
' 'A:c = 

. d V _ 



A?/ 



*"**> [2 ooj' + aAx]. § 6. 



Ax 



= 2 ax', or, in general, dy = 2axdx. § 17. 

ox 

By this general method we could differentiate any other func- 
tion, but in practice it is more expedient to use the rules which 
we proceed to establish. 



Algebraic Functions. 

19. The differential of the product of a constant and a vari- 
able is the product of the constant and the differential of the 
variable. 



ALGEBRAIC FUNCTIONS. 



11 



We are to prove that d (ay) = ady, in which y is some func- 
tion of x. Let u = ay, and let x' represent any value of x, and 
y ! and u' the corresponding values of y and u ; then 

u'=ay'. (1) 

Let Ax represent any increment of x, estimated from the value 
x\ and let Ay and Au represent the corresponding increments 
of y and u ; then 

u'+Au = a(y'-t-Ay) = ay'+aAy. (2) 

Subtracting (1) from (2), member from member, we have 

Au = a Ay. 



Ax 



Au 
Ax 
, limit VAu 

du dv 

. — = a— ■ 

dx dx 



limit 
Ax = 



Ay 

Ax 



( limit 



Ax 



§§ 6, 7, 
§ 17. 



Hence, as x' is any value of x, we have in general, by multi- 
plying both members by dx, 

du[=d (ay) ] = ady. 



a \a a 



20. The differential of a constant is zero. 
This is evident, since the increment of a constant in any 
interval of time is zero. 



21 . The differential of a polynomial is the algebraic sum of 
the differentials of its several terms. 

We are to prove that d (v+y— z+a) = dv + dy— dz, in which 
v, y, and z are functions of x. 

Let u = v-\-y — z-\-a, and let x' represent any value of x, 
and v\ y\ z\ and \0 the corresponding values of v, y, z, and u ; 
then 



^' + 2/'— z'-\-a. 



(i) 



12 



DIFFERENTIATION. 



Let Ax represent any increment of x, estimated from the value 
x', and Av, Ay, Az, and Aw the corresponding increments of #, 
y, z, and w ; then 

u' + Au = v'+Av+y'+Ay — (z r +Az) + a. (2) 

Subtracting (1) from (2) we have 

Aw = Aw + Aw — A3. 



Aw _ Av Aw Az ( 

Ax Ax Ax Ax 








limit rAw~j_ n m it 

Ax = Ax A.r = 


Aw Aw 
Ax Ax 


_Az~ 1 
Ax 


§ 6. 


du _dv dy dz 
dx dx dx dx 






§§ 10, 17. 



Hence, as x' is any value of x, we have in general 
du [ = d (v + y — z + a) ] == dw + % — cte. 



22. 27te differential of the product of two variables is the first 
into the differential of the second, plus the second into the differen- 
tial of the first. 

We are to prove that d (yz) = ydz + zdy, in which w and z are 
functions of x. 

Let w = yz, and let x' represent any value of x, and y\ z', and 
u' the corresponding values of w, z, and i£ ; then 

u'=y'z'. ■ . (1) 

Let Ax represent any increment of x estimated from the value 
x', and Aw, Az, and Aw the corresponding increments of w, z, 
and w ; then 

w' + Aw = (w' -f Aw) (z' + Az) 

= w 'z' + y' Az + z'Aw + AzAw. (2) 

Subtracting (1) from (2) we have 
Aw = y' Az + z'Aw + AzAw. 



Aw , 
Ax 



Az 
Ax 



+ («' + Az) 



Aj/ 
Ax 



ALGEBRAIC FUNCTIONS. 



13 



limit 
A;r = 



Au 
Ax 



limit 
Ax = 



fl Az 
Ax 



+ 



limit 
Ax = 



(s'+A*)^ 

Ax 



du ,dz , ,dy 
dx dx dx 



§§7,8. 



Hence, as x' is any value of aj, we have in general 
d« [= c?(yz)] = yc?» + zdy. 

To obtain this result geometrically, let 2 and ?/ represent the 
variable altitude and base of a rectangle 
conceived as generated by the side z moving h 
to the right, and the upper base y moving 
upward ; then zy = its area. 

If/ at the value dcba (Fig. 10) , dz — ah, 
and dy = ce, d(area) = cefb + bgha ; since 
cefb + bgha is evidently what would be the 
increment of the area of the rectangle in 
the assumed interval, if at the value dcba the increase of its 
area became uniform. 

Hence, d(zy) = d(area) = cefb + bgha = zdy + ydz. 

Here A (zy) = d (zy) + bgof. 







B 





Fig. 10. 



23. The differential of the product of any number of variables 
is the sum of the products of the differential of each into cdl the 
rest. 

We are to prove that d(xyz) = yzdx + xzdy + xydz, in which 
y and z are functions of x. 

Let u = xy, then d(xyz) — d(uz) . 

But d(uz) = zdu + udz § 22. 

= zd(xy) -\-xydz 

= yzdx + xzdy + xydz. 
.-. d(xyz) = yzdx + xzdy + xydz. 

In a similar manner, the theorem may be demonstrated for 
any number of variables. 



14 DIFFERENTIATION. 

24. The differential of a fraction is the denominator into 
the differential of the numerator, minus the numerator into the 
differential of the denominator, divided by the square of the 
denominator. 

We are to prove that d(K)— _ ? , in which y and z are 

\zj z\ 

functions of x. 

y 
Let u == '—•> then uz = y. 
z J 



. ' . udz + zdu = dy. 

dy — udz zdy — ydz 



dy — -dz 



,-.du = 



z z 



^ , /a\ xda — adx adx . 7 A ,, , . j7 

Lor. d I - = ^- = — — -, since da = ; that is, the 

\xj x 2, x z 

differential of a fraction with a constant numerator is minus the 
numerator into the differential of the denominator divided by 
the square of the denominator. 

25. The differential of a variable affected with any constant 
exponent is the product of the exponent, the variable with its expo- 
nent diminished by one, and the differential of the variable. 

I. When the exponent is a positive integer. 

If n is a positive integer, x n = x • x • x to n factors ; hence 
we have 

d(x n ) = d(x • x • x to n factors) 

= x n ~ Y dx + x n ~ l dx + etc. to n terms § 23. 

= nx n ~ 1 dx. 

II. When the exponent is a positive fraction. 

m 

Let y = fl5», then y n == .T m . (1) 

Differentiating (1), we obtain 
ny n ~ 1 dy = mx m ~ 1 dx. 



ALGEBRAIC FUNCTIONS. 15 



m x" 1 ' 1 7 m x m - l y _ m a m T a;» , 

,\ d?/ = ax— ax — ax 

* n y n ~ l n y u n x m 

= _# Ji dx, 
n 

III. When the exponent is negative. 

Let y = x~ 7 \ n being integral or fractional ; then 

'=?■ (1) 



Differentiating (1), we have 



m/y.W — 1 



eZy = —^-dx — — ?ia~' l_1 da. § 24, Cor. 



:ia 



For a proof of this theorem, which includes the case of in- 
commensurable exponents* see § 39, Ex. 25. 

Assuming the binomial theorem, let the student prove this 
rule by the general method of differentiation. 

Cor. d (Va) = \xr*dx = — — • 

2Va 

26. The general symbol for the differential off(x) is f'(x)dx ; 
hence, if y=f(x), dy =f'(x)dx. 



Examples. 
Differentiate 

1. x z + 8x + 2x 2 . Ans. (3a 2 + 8 +4 a?) dx. 

d (a 3 + 8x + 2 a 2 ) = d (a 3 ) + d (8a) + d (2a 2 ) , § 21. 

d(x 3 ) = 3x 2 dx, § 25. 

d(8x) = 8dx, § 19. 

d(2a 2 ) = 4ada, §§ 19, 25. 
.'. d (a 3 + 8a + 2a 2 ) = (3a 2 -f 8 + 4a) da. 

2. y=3aa? — 5wsc — 8m. cfa/ = (6aa — 5n)c?a. 

dy = d(3ax i — 5?ia— 8m) = c^aa 2 ) — cZ(5?ia) — d(8m). 
[The differentials of equals are equal, and § 21.] 



16 DIFFERENTIATION. 

3. f(x)= Sax 2 - 3b 2 x? - abx\ 

f'(x)dx= (10ax — 9 6 2 x 2 — 4a&x 3 ) dx. 

4. /(x) = a 3 + 5 b 2 x s + 7 a¥. 

/'(») da; = (15 6 2 x 2 + 35 a¥) dx. 

^ r i , ^ 3 ax + b , 

5. 2/ = ax§ + 6x* + c. dy = ^-ax. 

2Vx 

6. y 2 =2px. ty—P. 

d{y 2 ) — d{2px). dx~ y 

7. oV + 6 V = a 2 b 2 . dy=-^ dx. 

8. f(x) = {b + ax 2 )*. f'(x)dx = %(b + ax 2 )iaxdx. 

9. ? / = (l+2x 2 )(l+4x 3 ). rtj/ = 4x(l+3x + 10x 3 )ax. 

dy = (I + 2x 2 )d(l + 4x s ) + (1 +4x 3 )a'(l + 2x 2 ). 

. A x + a 2 , 6 — a 2 , 

10. ?/ = — ! dy = -dx. 

y x + b . y (x + 6) 2 

dy=i (x + b)d{x + a 2 )-(x + a 2 )d{x + b) 



{x + by 

—• f'(x)dx = — — . . 

b-2x* J K J (b-2x 2 )'' 



11- /(*)=— ^ f(x)dx= ^ * a * <fa 



a — 3x 



12. ?/ = (a+ x) Va — x. dy = dx 

2Va — x 

,o ^/ x 2x A ,/s \j 8a¥- 4x 5 , 



a 2 — x 2 (a 2 — x 2 ) 






dx 

m+l 



(l-x)Vl-x 

16 -^)=(T^i- ^ = (1^)1 

17. v = — • 

Vi+* 2 



DERIVATIVES . 17 

3 a? _ 2 y 2 

18. 2xy 2 — oy 2 = or 3 . dy = — *— dx. 

4xy — 2ay 

19 . m =¥±± f(x)dx= ^+^+v dx . 



±x + a? v ' {-ix + x 2 )' 



20. y = 



or — ar 



21. f{x)=^/ax + -\/&f. f'(x) ^ Va " 



2V.r 

! < 



ao a 6 ax 7 

22. ^- — •• dx. 



23. /(») = — -■ /'(a)= 3a;2 + a;8 . 

/v y (1+a) 2 J w (l + ») 3 

24. /(a^-M^L. /'(«) = l +±* 



25. (* ~ 9y) (ft -te) = («-*) (!-*)■ 

dy _ b — 4 x + 6 y — 2 ay 
cfee a 2 — 6 # — e + 26 

27. An Increasing function is one that increases when its 
variable increases ; hence it decreases when its variable de- 
creases. 

A Decreasing function is one that decreases when its variable 
increases ; hence it increases when its variable decreases. 

Thus, ax and a x are increasing, and - and a — x are decreasing 

x 
functions of x. 

28. The Derivative of a function is the ratio of the differen- 
tial of the function to the differential of its variable. This ratio 
is sometimes called the derived function or the differential coeffi- 
cient. Hence the derivative of f{x) is /'(a), or the ratio of 
f'(x)dx to dx. The derivative of y with respect to x is repre- 
sented by ^. If y =f(x) , then c ll = f(x) . Thus, if y= x 3 , 

j " dx dx 

— =3x 2 ; that is, 3 x 2 is the derivative of ?/, or x J ; if f(x) = .t 6 , 
then f(x) = 6ar\ 



18 DIFFERENTIATION. 

29. The Measure of the rate of change of a variable at a 
given instant is what ivould be its increment in a unit of time, 
if at that instant its change became uniform. This measure 
of rate is generally called the rate. Hence, the rate will be 
positive or negative, according as the variable is increasing 
or decreasing. Thus, when we say that the distance of a train 
from the station was changing, at a given instant, at the rate 
of -+-30 miles an hour, we mean that this distance would have 
increased thirty miles in an hour, if at that instant its increase 
had become uniform, 

If the change of a variable is uniform, the actual increment 
of the variable in a unit of time is the measure of its rate. 

30. Signification of — . Let t represent time ; then, any vari- 
able, as y, is evidently some function of t. Since time changes 
uniformly, dt may represent any increment or interval of time. 
If dt equals the unit of time, then by definition cly equals the 
measure of the rate of change of y ; and, if dt is n times the unit 
of time, dy is n times the rate of change of y ; hence, whatever 
be the value of dt, — = the rate of change ofy. 

31 . Signification of ^, or /' (x) . c ]l = a l^ a ^ == the ratio 

dx clx dt dt 

of the rate of change of y to that of x (§ 30). f'(x) = / > ' 1 

at az 

= the ratio of the rate of change of f(x) to that of x. 

Hence, the derivative of a function expresses the ratio of the rate 
of change of the function to that of its variable; and a function 
is an increasing function or a decreasing function, according as 
its derivative is positive or negative. 

Cor. The same function of x may be an increasing function 
for some values of x, and a decreasing one for other values. 

Thus, since -, the derivative of — , is -f when x < 0, and 

x 3 ar 

— when x > 0, — is an increasing function when x<0, and 

x 2 
a decreasing function when x > 0. 



APPLICATIONS. 19 

32. When the change of y is uniform, it is evident that — 
6 J At 

is the rate of change of y. When the change of y is variable, 
the value of — - evidently lies between the greatest and the least 
values of the rate of change of?/ during the time At ; hence, the 

Ay 

smaller At is taken, the nearer — approaches the rate of change 
of y at the beginning of At. 



Hence limit 
.rience, A ^ Q 



At 



the rate of change of y \ _ dy 
at the beginning of At ) dt 



~„a limit 

and A ^ Q 



Ax __ j the rate of change of x } _dx 
At _ I at the beginning of At) dt 

Dividing (1) by (2), we obtain, without the aid of a locus, 



(i) 

(2) 



limit 
Ax=0 



Ax_ dx 

Applications. 

1. The area of a circular plate of metal expanded by heat 
increases how man}' times as fast as its radius ? If, when the 
radius is two inches, it is increasing at the rate of .01 inch a 
second, how fast is the area increasing at the same time ? 

Let x — the radius, and y the area of the plate ; then y = ttx 2 . 

.'. dy = 2irxdx, or -^- = 2 ttx— ; that is, the area is increas- 
ed dt 

ing in square inches 2ttx times as fast as the radius is in linear 

dx dv 

inches. When x = 2, and — = .01 in., -^= .04 -w sq. in. ; that 

dt dt L 

is, the area is increasing .04 tt sq. in. a second at the instant 
considered. 

2. The volume of a spherical soap-bubble increases how many 
times as fast as its radius? When its radius is 3 in., and is 
increasing at the rate of 2 in. a second, how fast is its volume 
increasing ? 

Ans. The volume is increasing in cubic inches 4 ttx 2, times as 
fast as the radius is in linear inches. The volume is increasing 
72-7T cu. in. a second at the instant considered. 



20 DIFFERENTIATION. 

3. A bo}~ is running on a horizontal plane in a straight line 
towards the base of a tower 50 metres in height. He is 
approaching the top how many times as fast as he is the foot of 
the tower? How fast is he approaching the top, when he is 500 
metres from the foot, and running at the rate of 200 metres a 
minute ? 

Let x and y respectively represent in metres the distances 
of the boy from the foot and the top of the tower ; then 
y 2 = x 2 -f- (50) 2 , etc. Ans. 199 metres a minute. 

4. A light is 4 metres above and directly over a straight 
horizontal side-walk, on which a man If metres in height is 
walking away from the light. The farthest point of the man's 
shadow is moving how many times as fast as he is walking? 
The man's shadow is lengthening how many times as fast as he is 
walking? How fast is the shadow lengthening, and its farthest 

point moving, when the man is walking 
at the rate of 50 metres a minute ? 

Let ae be the sidewalk, b the position 
of the light, and cd one position of the 
man. Let ae = ?/, and ac = x ; then 
y — x : y : : % : 4 ; . ' . dy — 1 ^-- dx. Again, 
let 2/ = CE, and o? = ac; then y + x :y : :4 : f ; .\dy = j-dx. 

5. The altitude of a variable cylinder is constantly equal to 
the diameter of its base. In general, its volume is changing how 
many times as fast as its altitude ? If, when its altitude is 6 
metres, it is increasing at the rate of 2 metres an hour, how 
fast is its volume increasing at the same instant? How fast is 
the entire surface increasing at the same instant? 

Ans. %ttx 2 times, x being its altitude; 54 ?r kilolitres an 
hour ; 36 tt centiares an hour. 

6. The altitude of a varying frustum of a right cone is con- 
stantly equal to the radius of its lower base, and the radius of 
its upper base is one-half that of its lower base. If, when the 
radius of its lower base is 4 metres, it is increasing at the rate 
of 2 metres an hour, how fast is the volume of the frustum 
increasing at the same instant? 




APPLICATIONS. 21 

7. The area of an equilateral triangle increases hew many 
times as fast as each of its sides ? How fast is its area increas- 
ing when each of its sides is 10 in., and increasing at the rate 
of 3 in. a second? What is the length of each of its sides, when 
its area is increasing in square inches 30 times as fast as each 
of its sides is in linear inches ? 

Ans. 15 Vo sq. in. a second; 20V3 in. 

8. One end of a ladder 20 ft. long was on the ground 5 ft. 
from the foundation of a building, which stood on a horizontal 
plane, while the other end rested against the side of the build- 
ing. The end on the ground was carried away from the build- 
ing on a line perpendicular to it, at the uniform rate of 4 ft. a 
minute ; how fast did the other end begin to descend along the 
building? How fast was it descending at the end of two 
minutes? How far was the foot of the ladder from the building, 
when the top was descending at the rate of 4 ft. a minute ? 

Ans. 1.03+ ft. a minute ; 3.42 ft. a minute ; 10 V2 ft. 

9. In the parabola whose parameter is 8, the ordinate 
changes how many times as fast as the abscissa? What is its 
slope at any point (x, y) ? Find its inclination at the points 
whose common abscissa is ^. Is y an increasing or a decreas- 
ing function of x? At what points does the ordinate change 
numerically four times as fast as the abscissa ? 

4 
In this case, y is a two-valued function; aud - is + or — , 

y 

according as y is + or — ; .". the + value of?/ is an increasing, 
and the — value a decreasing, function of x. 

Ans. -; t; 63° 26' 6" and 116° 33' 54"; a, 1) and (1,-1). 

y y 

10. In the ellipse cry 2 + b 2 x 2 = a 2 b 2 , the ordinate increases 
how many times as fast as the abscissa ? y changes how 
many times as fast as x at the extremities of the axes of the 
curve ? How can the points be found at which y changes c times 
as fast as x? What is the slope of the ellipse at any point? 
What, at the extremities of its axes? Is y an increasing or a 
decreasing function of x? 



22 DIFFERENTIATION. 

dy b 2 x . , , ,. „ ' & 2 a? 

-^■= — ; .". when y changes c times as last as a?, — = c. 

da? a 2 ?/ , 2 a-?/ 

When x and 2/ have unlike signs, — is + , and 2/ is an 

increasing function ; when x and y have like signs, — is — , 

and y is a decreasing function. 

11. What is the slope of y 2 = X s + 2 x 4 at (#, y) ? What is it 

Ans. ± X + - .. ± i9.VlO. 

2Va+2z 2 10 

12. What is the slope of ?/ = X s — x 2 -f- 1 at the point whose 
abscissa is 2? 1? 0? -1? Ans. +8; +1; +0; +5. 

13. At what point on y 2 = 2 X s is the slope 3 ? At what point 
is the curve parallel to the axis of x? Ans. (2, 4) • (0, 0). 

14. At what angles does the line Sy — 2x— 8 = cut the 
parabola y 2 = Sx? 

Find their slopes at their points of intersection ; then find the 
angles between the lines having these slopes. 

Ans. tan -1 . 2 and tan -1 . 125. 

15. One ship was sailing south at the rate of 6 miles an hour ; 
another, east at the rate of 8 miles an hour. At 4 p.m. the 
second crossed the track of the first at a point where the first 
was two hours before. How was the distance between the ships 
changing at 3 p.m. ? How at 5 p.m. ? When was the distance 
between them not changing? 

Let t = the time in hours, reckoned from 4 p.m., time after 
4 p.m. being +, and time before — . Then 8t and 6£+ 12 will 
represent respectively the distances of the two ships from the 
point of intersection of their paths, distances south and east 
being +, and distances west and north being — . Let y = the 
distance between the ships ; then, 

2/ 2 = (80 2 + (6^ + 12) 2 . 

' ' dt ~[>U 2 + (6£ + 12) 2 ]i' ^ ) 



VELOCJTY AND ACCELERATION. 28 

When y does not change, dy=0. /.from (1), 100 £ + 72 = 0; 
.-. t= — .72 of 60 minutes =— 43.2 minutes. Therefore, the 
distance between them was not changing at 43.2 minutes before 
4 p.m., or at 16.8 minutes after 3 p.m. 

Ans. Diminishing 2.8 miles an hour; increasing 8.73. 

33. Velocity is the rate of change of the distance passed over 
b}~ a moving body. Hence, if s = the distance and v = the 

velocity, v = — (§ 30). If the unit of s is one foot, and the 

ds 
unit of t one second, v = — ft. a second. 
dt 

Acceleration is the rate of change of velocity. 

dv 
Hence, if a = acceleration, a = — (§ 30). 



Examples. 



1 . If s = 2 f, what is the velocity and acceleration ? 

Here v = — = 6t 2 ft. a second ; a = — = 12£ ft. a second ; 
dt dt 

and the rate of change of acceleration = — = 12 ft. a second. 

dt 

2. If g = 32.17 ft., s =j-t 2 is the law of falling bodies in 

vacuo near the earth's surface ; find the velocity and accelera- 
tion in general, also at the end of the third and the eighth 
second. 

Ans. a = 32.17 ft. a second ; v = 32.172 ft. a second ; 96.51 ; 
257.36. 

3. Given s = aft 9 to find v and a in general, and at the end 
of 4 seconds. 

Ans. v = and - ft. a second ; a = and ft. 

2 V* 4 Wf 32 

a second ; that is, the A^elocity decreases at the rate of ft. 

Wt 3 
a second. 



24 DIFFERENTIATION. 

4. Given s 3 = 8t 2 , to find v and a in general, and at the end 
of 8 seconds. 

Ans. v = — - — and f ft. a second. 
3-y/t 

5. A point moves along a parabola with a velocity v' ; re- 
quired the rates of change of its coordinates. 

Since y 2 — 2px, _^=-£-. (\\ 

dx y v y 

If s represents the length of the curve traversed, by the 
conditions of the problem, we have 

*="' (2) 

But ^ S •— ^ - ^ s — ^ - ^d°(? + dy 2 _ dy L.dx^ .„. 

dt dt ' dy~dt' dy ~dt\ dy 2 ' ^ 

since ds = -Vdx 2 -f- dy 2 . § 16, Cor. 2. 

from (1), (2), and (3), we obtain 

dt \ p ^ Vp 2 -f2/ 2 

which is the rate of change of y. 

In like manner, we obtain 
dx _ y 



dt 



V/r -f y- 



v\ the rate of change of x. 



Logarithmic and Exponential Functions. 

34. The differential of the logarithm of a variable is the quo- 
tient of the differential of the variable divided by the variable 
itself, multiplied by a constant. 

Let y = nx; (1) 

then dy = ndx, (2) 

and \og a y*= \og a ii + \og a x. (3) 

* log a y is read, "the logarithm of y to the base a" 



LOGARITHMIC AND EXPONENTIAL FUNCTIONS. 25 



(4) 



From (1) and (2), 

dy dx 

V x 
From (3), 

d(\og a y) = d(log a x). (5) 

From (5) and (4), 

d(\og a y) d(\og a x) 
dy dx 

~y ? 

dx 
Whence d(\og a x) bears the same ratio to — that d(\og a y) 

does to JL. Let m be this ratio for some particular value of a?, 

y 

as x 1 ; then d(log a #) = m — when a? = x\ and d(log ffl ?/) = m _~ 

when y = nx' ; but, as w is an arbitrary constant, nx' may be 

any number. Hence, in general, d(\og n y) = m — , in which m 

2/ 
is a constant.* 

The constant m is called the Modulus of the system of loga- 
rithms whose base is a. 

35. Let m and m 1 be the moduli of two systems of loga- 
rithms whose bases are a and b respectively. If a > 6, 
it is evident that \og a x must change more slowly than log b x; 
• '. d(\og a x) < d i\og h x) ; that is, 

dx ' .dx , 

m — < m' — , or m < m\ 

Hence, the greater the base of a system of logarithms, the smaller 
is its modulus. 

36. Naperian System. The system of logarithms whose 
modulus is unity is called the Naperian system. The symbol 
for the Naperian base is e. 

* See Rice and Johnson's Calculus, p. 39 ; Olney's Calculus, p. 25 ; also 
Bowser's Calculus, p. 29. 



26 DIFFERENTIATION. 

The differential of the Naperian logarithm of a variable is, 
therefore, the differential of the variable divided by the variable. 

Thus we see that Naperian logarithms are the simplest and 
most natural for analytic purposes ; and, hereafter, the symbol 
log will stand for the Naperian logarithm. 

37. The differential of an exponential function with a constant 
base is equal to the function itself into the logarithm of the base 
into the differential of the exponent, divided by the modulus of the 
system of logarithms used. 

Let y = c x , then \og a y = x log a c ; 

dy c x \og a c 
.'.m — = log a cc£r ; .'. dy\=d(c x )l = dx. 

Cor. In the Naperian system, the modulus being unity, we 

have 

d(c x ) = c x \ogcdx-, 

also, d(e x ) = e x dx, since loge = 1. 

38. The differential of an exponential function with a variable 
base is the sum of the residts obtained by first differentiating as 
though the base were constant, and then as though the exponent 
were constant. 

Let u — y x , then log u = x logy ; 

du , , dy 

. = \ogydx -f- x— ; 

u & ^ y 

:, du = y x \ogydx -f xy x ~ l dy, 
which is the result obtained by following the rule given. 

39. Logarithmic Differentiation. Exponential functions, as 
also those involving products and quotients, are often more 
easily differentiated by first passing to logarithms. This method, 
which is illustrated in the two preceding demonstrations, is called 
logarithmic differentiation. 



LOGARITHMIC AND EXPONENTIAL FUNCTIONS. 27 



Examples. 



1. y = \og{x i -\-x ) 



2. y = \og a -Vl—x\ 

3. f(x)=\og a x\ 

4. f(x) = x log x. 

5. y = c Xosax . 

6. /(*) = (log a) 3 . 

7. /(*)=** 

8. y=-a^. 



dy = - — i— dx. 
x 2 -\-x 

j 3 mx 2 
ay = — — ; : ax. 



/'(*) 



2(^-1) 
3 m 



f'(x) = logx+l. 



dy = 



logc 



dx. 



/'(tt)cfa= 3(logg)2 .fa. 

X 

f'(x) = x?(logx + l). 



logy = x x log x; .*. ~ = x x — + log a? [af (log a; -f- l)]da?o 
logx(loga;+l) + 



,\dy = x? x* 



x 



dx. 



9 . ?/ = a?**. 

a 2 — 1 



% _ x e* e x 1 4-^logx 
dx X 



10. 2/ 



a z + l 



d*/ 



2 a 1 log ado; 



11. y = log 



VlT^ 
12. ?/ = log (log x). 



13. 2/ = — • 



(«* + l) 2 
dx x(l 4-ic 2 ) 
C?iC x log X 

da; ~~ [xj \ ° x J 



14. y = 



Vl+x 



VT-a; 



dy 



dx 



(l-^)Vi-^ 2 



In this example and some that follow, pass to logarithms. 



28 








15. 


y = 




x n 


(a 


+ x) n 


16. 


y = 


(a 


2 + z 2 )§ 



DIFFERENTIATION. 



dy = 
dx 


1 
1 + e* 


dy _ 


x x (1 — logo?) 


dx 


X 2 


dy = 
dx 


= e x (l-3x 2 -x"). 


dy _ 


4 


dx 


(e x + e~ x y 


dy = 

dx 


='iT( 1+log f 



dy _ nax n ~ l 
. dx~ (a + x) n+1 

^ = 2a? 2a '-^ (a 2 + a?)K 
v^u, — jo )* dx (a 2 — ar)i 

17. ?/ = log 

J S l + e* 

i 

18. 2/ = of, 

19. y = e x (l-x 3 ). 

20. y = ^£^ 

«■ »-$r- 

22. 2/=(a a: +l) 2 . dy = 2a x (a x + l)logadaj. 

23. Which increases the more rapidly, a number or its 
logarithm ? 

Let 2/ = log a a?, then dy = —dx\ hence log a # changes faster or 

x 

more slowly than #, according as»< or > m. 

Since, in the Naperian sj'stem, m = 1, dx = xc??/ ; that is, the 
number a? changes x times as fast as log e a\ 

Hem. The ratio of the rate of change of a number to that of 
its logarithm is variable ; and yet the hypothesis, that it is con- 
stant for comparatively small changes in the number, is 
sufficiently accurate for practical purposes, and is the assump- 
tion made in using the tabular differences in tables of logarithms. 

24. If y = log 10 x, x changes how many times as fast as ?/, 
when x = 2560, the modulus of the Common system being 

.434294? 

da = x = 2560 =5895nearl 
dy ra .434294 J 



TRIGONOMETRIC FUNCTIONS. 29 

dx 

25. By means of the formula d(\ogx) — — , find d (x n ) in 

which n is any number, commensurable or incommensurable. 
Let u = x n , then log u = n log x ; 

dit dx , __! , 

.*. — = n — , or du = nx n l clx. 
u x 

If x were negative, to avoid logarithms of negative num- 
bers, we would square both members of u = x n before differen- 
tiating. 

26. In like manner, obtain d(xy), d(xyz), and dl-y 

27. Prove that d(— log — - + logc) = 



2 a " x + a y or — a - 

d07 



28. Prove that d [log (a +Vr ± a 2 ) -f log c] = 



vrift 2 



29. What is the slope of the curve x = log 1Q y, or y = 10 T ? 
What at x = 0? What at y = 5? 

J. ?ls . J — • 2.3+; 11.51 + . 
.43429 



Trigonometric Functions. 

40. In the higher mathematics, the tm& of angular measure 
is the angle whose measuring arc is a radius in length ; hence, 
if x represents the length of the measuring arc of any angle, 

x 
and r its radius, the angle equals - ; or, if r = 1, the angle = x. 

r 

In what follows we shall assume r=l. 



41. Tfie differential of the sine of an angle is equal to the 
cosine of the angle into the differential of the angle. 

TJie differential of the cosine of an angle is equal to minus the 
sine of the angle into the differential of the angle. 



30 



DIFFERENTIATION. 




For, let x represent any angle, or its measuring arc, and let 
ac be any value of this arc. If at c the mo- 
tion of the generatrix became uniform along 
the tangent cd, it is evident that any simul- 
taneous increments of its distances from c 
and lines ba and bh may be taken as the 
differentials of the arc, the sine, and the 
cosine, when x = ac ; that is, if cd = dx, 
ed = d(smsc), and — ec = cl (oosx). Now 
angle edc = abc = x ; .'.in triangle edc, 
ed [ = d (sin x) ] = cos x dx, 
and — ec[= d(cos x)~\ = — sin xdx. 

Hence, as abc is any value of a?, we have in general 
dsmx — cos xdx, 
and d cos x= — sin xdx. 

42. The differential of the tangent of an angle is equal to the 
square of the secant of the angle into the differential of the angle. 
sin x 



For 



tan& = 



.'.dtana; 



cos# 
cos^dsinx 



sinit'dcoscc 



(cos 2 # + &m 2 x)dx 

COS 2 £ 

dx 



= sec 2 xdx. 



cos-o? 



43. TJie differential of the cotangent of an angle is equal to 
minus the square of the cosecant of the angle into the differential 
of the angle. 



For cot x = tan f - — x 



.'.dcotx = sec 2 f — — x \d( —x ) = 



■cosec 2 xdx. §42. 



44. The differential of the secant of an angle is equal to the 
secant of the angle into the tangent of the angle into the differential 
of the angle. 



For 



sec a? = 



.'.dsecx = 



TRIGONOMETRIC FUNCTIONS. 

1 
133 

sinxdx 



31 



cos a 

dcosx 



= secxt&nxdx. 



cos~# 



COS z £ 



45. The differential of the cosecant of an angle is equal to 
minus the cosecant of the angle into the cotangent of the angle into 
the differential of the angle. 



For 



cosec x 



= secf - — x ] ; 

.'.d cosec x = sec ( - — # Jtanf- — x \dl - — x\ § 44. 
= — cosec x cot a? da?. 

46. d vers x = d(l — cos x) = sinxdx. 

47. d* covers x = d(l — sin sc) = —cos £ecfcc. 

48. To prove these theorems by the method of limits, we 
need the following lemma, which is very useful also in the 
theory of curves. 

Lemma. The limit of the ratio of an arc of any plane curve 
to its chord is unity. 

If s represents the length of 
the curve mw, and pb = dx, 
pd = ds ; and if pc = Aaj, arc 
pap' = As. Since s is a function 
of x, we have 

limit |~ Ag ~| = ds - 
A * = °|_Aa,-J dx 

But limit [" chord ppH ds 

A * = °|_ Ax 




dx 



Fig. 13. 



limit 



since . A Tsec cpp'1 = sec bpd. 



Hence, by division, we have 



limit 
As = 



A s 

chord 



— 1 = 

:d pp'J 



1. 



§9 



32 



DIFFERENTIATION. 



Cor. Since one-half of the chord of an arc whose radius is 
unity is the sine of half the arc, 



limit 
x^=0 



1. 



49. To prove that dsin x = cos xdx by the method of limits. 

Let y = sin x ; 

then Ay = sin (x -f- Ax) — sin a?. 

But, from Trigonometry, we have 

sin a? — sin?/ = 2cos-j-(# -\-.y) sm^-(# — y). 

.'. Ay = 2 cos (a? -f -J Aa?) sin \ Ax. 

Ay , , , A x sinAAa; 
.-. -^- = cos(# -f- f Ax) * — ■ • 



Ax 



i-Ax 



Now 
and 



[im lo [cos(# -f- ^ Ax)~] = cos a; ; 



Ax 
limit 



n j- AaT j 



sin ^ Aa?~ 



1. 



§ 48, Cor 



dy 

.'.-?- = cosrc. 
dx 

The other theorems can be proved in like manner. 



Examples. 



Differentiate : 

1. sin ace. 

2. ?/ = cos-. 

a 

3. y — cos X s = cos (cc 3 ) . 

4. /(a) = tan m a = (tanjc)' 

5. /(#) = tan x -f- sec cc. 

6. y = sin (log x) . 



^4?is. acos^da;. 

dv 1 . as 

-^-= sin-- 

dec a a 

dy= —3x 2 smx i dx. 
f'(x)dx = m tan TO-1 # sec 2 a; da; 



/'(*)= 



1 + sin x 



COS - £ 

d?/ cos (log x) 
dx~~ x 



7. y — log (tana?) 



TRIGONOMETRIC FUNCTIONS. 33 

dy 2 2 



dx 2 sin a; cos x sin 2 x 



8. y = log(sinx). — = cota\ 

Of a? 

9. y = log(cotcc). -^ = 



da? sin 2 # 

10. y= 1 ~ tana? [=cosa;-sina;]. 

seax 

11. y = £c n e sina: . cZy = a5 n_1 e 8ina: (w 4- a? cos x)dx. 

12. ?/ = sin-(?ia;)sm n a,*. dy = nsm n ~ 1 x sin (nx-{-x)dx. 

13. y = e x log sin a;. dy = e x (cot x + log sin a;) dx. 

14. ?/ = tan (log x) . 

15. ?/ = logsec#. 

1 ^ cos flJ 

2 sin 2 a; 

17. i/ = 4 sin m ace. dy = 4 am sin m_I aa; cos «# da?. 

18. y = af in *. ^af*"/— + log x cos a 

da? \ x & 

19. y = (sina?) tana: . — = (sina;) tanz (l-f-sec 2 a;logsina?). 

CiX 



J_ 1 

,2, ~ 



20. y = tana z 



dy_ a J sec 2 a x log a 



->.. tan 3 « , . dy , 4 

21. y = tan# + a,\ -^ = tan 4 #. 

^3 da; 

22. y = e (a+I ' 2 sina;. ^ = e (a+x)2 [2(a+tf)sina-f cos a;]. 

23. y = e ax ^ cos nc. -*■ = — e (2 crx* cos ?*a; -f- r sin ra;) . 

dx 



^. , /^ dy — (secv 1 — #) 2 

24. y = tanvl — a;. -£ = — ^ , — *—• 

dx 2Vl-a? 



34 DIFFERENTIATION. 

25. Are sinx, cos a?, and tana? increasing or decreasing func- 
tions of x? 

d cos x = — sin x dx ; and — sin x is positive when x is of the 
third or the fourth quadrant, and negative when x is of the first 
or the second ; hence, cos a? is an increasing function when x is of 
the third or the fourth quadrant, and a decreasing function when 
x is of the first or the second, d tan x = &ec 2 xdx, and sec 2 a; is 
always positive ; hence tan x is an increasing function of x 

between its limits of continuity ; that is, between x= — - and 

x = -, etc. 
2 

26. At what values of x does sin x change as fast as #? At 
what values does cos x? tan x? cot x? 

Ans. sin x does, when x =• and 7r ; 

cos x does, when x = - and f w. 

2 2 

27. If the change of x and cos a* became uniform at 30°, 
how much would cosx decrease while x increases from 30° to 
30° 15'? 

Let y = cos x ; then dy = — sin xdx — — Jcfa, when x = 30°. 

Let dx = 15' = 3,U159 = .004363 ; then cfy = - .002182. 
180x4 

Hence cos x would decrease .002182. This is evidently less 
than the actual decrement. 

28. A vertical wheel whose circumference is 20 ft. makes 5 
revolutions a second about a fixed axis. How fast is a point 
in its circumference moving horizontally, when it is 30° from 
either extremity of the horizontal diameter? 

Ans. 50 ft. a second. 

29. What is the slope of the curve y = sinx? Its inclination 
lies between what values ? What is its inclination at x = ? 

What at » = -? 
2 
The slope = cos x ; hence, at any point, it must be something 
between —1 and +1 inclusive. Hence, the inclination of the 



ANTI-TRIGONOMETRIC FUNCTIONS. 35 



curve at any point is something" between and -, or something 
between f?r and it inclusive. 

Ans. -: 0. 



30. What is the slope of the curve y = tan x? Its inclination 
lies between what values ? What is its inclination at x = 0? 

Whatat<B = -? 

4 

Ans. see 2 #; between - and - inclusive ; -; 63° 26' 6"- 
4 2 4 



Anti-Trigonometric Functions. 
clx 



50. ^(sin- 1 ^) = 



"V 1 — XT 

Let y = sin _1 ic, then x = sin y ; 



.*. dx = cosycly = Vl — siirydy = Vl — a^cfa/. 

/. d?/[=cZ(sin- 1 ^)]= ^ • 

VI- a? 

( 

Vi^ 



51. ^(cos- 1 ^) = d(-- sm *aA = - jf_ - §50. 



52. d(tan- 1 o,')= ' 



Let y = tan -1 #, then x = tan ?/ ; 

.-. dx = sec 2 ydy = (1 -f- tsm 2 y)dy — (1 -f- a^cfa/. 

.•.%[=d(tan- 1 a;)]=— fL- 

* To avoid the ambiguity of the double sign ±, we shall, in these formu- 
las, limit sin- 1 ^, cos -1 x, etc., to values between and -. They maybe 

made general, however, by writing the double sign in the second member 
of each. 



36 DIFFERENTIATION. 

53. d(cot- 1 x)=d(l-tsin- 1 x) = 

54. d(sec~ 1 x) = 



2 J l + x 2 

dx 



xv x 2 — 1 
Let y == sec" ' 1 aj, then a? = sec ?/ ; 



.*. da? = sec y tan ydy = xvx 2 — 1 cfa/. 
'.•. dy [= d(sec -1 ^)] = 



a^Va? 2 — i 

die 



55. d(cosec" 1 a?) = d( - — sec _1 ic ] = — 

56. d(vers" _1 aj) = 



.2 / a^V^ 2 — 1 

da; 



V2a> — a? 2 
Let y = vers -1 a; , then a; = vers ?/ ; 



/. dx = smydy = Vl — cos 2 ?/ d?/ 

= Vl — (1 — vers y) 2 dy 



= V 2 vers y — vers 2 ?/ d?/ 



= V 2 x — x 2 dy. 

/. d?/ [ = d (vers -1 a?)] = — • 

V2 x — x 2 



57. d (covers" 1 #) = d f| — 



vers -1 x i = — 



dec 



V2# — x 2 



1. Prove that 



Examples. 
dx 



d ( sin "1)= 

I sin - , 

V a J . U A'V Va 2 -x* 



Va 2 — a? 2 
aA W da; 



v-e 



50. 



ANTI-TRIGONOMETRIC FUNCTIONS. 37 

2. Prove that d( cos 1 - )= X ; dl tan -1 - ) = -r— — a \ 

>*) = ^^; d{sec-i*)= adx 



d cot" 1 - =- 



a + #" V <V War* — a 

,/ iJcN ada^ ,/ _!^\ dx 

dl cosec- — ; a[ ™« - 1 - > — 



\ a J x Va^ — a 2 V c v "\Z2ax-x 2 

-,{ -\%\ dx 

dl covers - = • 

V a / -\Z2ax-x 2 



3. ?/ = a? sin 1 cc. — ^=sin _1 aH • 

^ Vl-arv 

j 4._i ^V b 4. -i i tana; 

4. y = tan a; tan 1 a;. -# = sec a; tan" \t H -• 

dx 1 H- af 

. -i 2a; 
o. ?/ = tan x -« 

~ . _!a;+l 

6. y — sin 1 — ■ — 



-i 1 
/. y = sec ■ 



2ar 2 -l 



w = cos * 

ar 9 » + 1 



9 . y = tan * (?i tan a;) . 



^2/_ 


2(l-ar*) 


cfcc 


l + 6ar 9 + a; 4 


*/_ 


1 


da; 


Vl-2a;-a^ 


<fy_ 


2 


da; 


Vl-ar* 


dx 


2 na; n_1 
ar^ + 1 


d y _ 


?i 



da; cos" x -f- ?i- sin- a; 



10 . y^a-.-V. ^ = afin -i,/sio_^ + _Jo^_Y 

da; V x Vl-arV 

11. A wheel whose radius is r rolls along a horizontal line 
with a velocity i>' ; required the velocity of any point p in 
its circumference ; also the velocity of p horizontally and 
vertically. 



38 DIFFERENTIATION. 

Let apx (Fig. 14) represent the cycloid traced by the point p, 

referred to the axes ax and 
T ~ •£^T < \ ay ; then will the horizontal 

and vertical velocities of p 

be the rates of change of x 

and y respectively, d be- 

x ing the point of contact, 

ad = r vers *^- Since the 
r 

centre is vertically over d, its velocity is equal to the rate of 
increase of ad ; 

... v' = ^Yrvers^ = r S. 

dt \ rj -^2ry-y 2 M 




~ = y. — 9-v' = the velocity vertically. (1) 



Since ae = ad — ph, and ph = -\A/(2r — y) , the equation of 
the cycloid is evidently 



^ 



x = r vers 1 V2 ry — y 2 . 

,\ dx = ^ dy. 

■y/2ry-y 2 

Dividing by dt. and substituting the value of -A we have 

d x v dy v , * , . ,i 

— = ■ - — = -v f = the velocity horizontally. (2) 

tit V 2 ry — y °^ r 

From § 16, Cor. 2, and equations (1) and (2), we have 

ds = ^/dtf + dy 2 = ( 2ry 7 7 2/8 + pjv'dt =yj^v'dt. 

ds (2v 
.'.— =^-f v' = the velocity of p ; (3) 

for the velocity of p equals the rate of increase of s. 

* The symbol — indicates the operation of taking the derivative with 
respect to t. 



MISCELLANEOUS EXAMPLES. 



39 



if 



if 



if 



From (1), (2), and (3), we have, 
dy n dx 



y = o, 



dt 



0, 



dt 



ds 
0, and — = : 

dt 



y = i 



dy , dx , , ds rx , 



dt 



y = 2 r, -^ = 0, 



(ft 



dt 

ds 



= 2v , and — = 2v'. 
dt dt 



Hence, when a point of the circumference is in contact with 
the line, its velocit}' is zero ; when it is in the same horizontal 
plane as the centre, its velocity horizontally and vertically is the 
same as the velocity of the centre ; and when it is at the highest 
point, its motion is entirely horizontal, and its velocity is twice 
that of the centre. 



Since 



ds 
dt 



-j?- 



V2 



— v' 



. ds , /t: — 
. . — : v : : v2ry : r. 
dt y 

Hence the velocity of p is to that of c as the chord dp is 
to the radius dc ; that is, p and c are momentarily moving 
about d with equal angular velocities. 



Miscellaneous Examples. 
1. If y =/(#), show that 



limit 
Ax=0 



Aa; 



limit 



f(x + Ax)-f(x) 
Aa; 



2. Find — ( 



= f(x). 

Ans. 



3. Find 



dx 



.O-^) 1 . 



(a 2 -x 2 )i 

Sx 2 
(l-^)f' 



40 DIFFERENTIATION. 

4. f(x) = (x -3)e 2x + 4xe x + x + 3. 

f'(x) = (2x-b)e 2 * + 4:(x+l)e x +l. 

5. y = log tan" 1 x. $& = L___. 

dx (1 +aj 2 )tan" 1 aj 

6. Find— [a^-SlogCl + aj 8 )*]. 



dx 1 + x J 

x loaf x , , /1 v dy loarx* 

■ 1-x BX } dx (l-x) 2 



_ y £ 2 + a 2 + V^+T 2 



Vcc 2 + a 2 - Vx 2 + & 

dy _ 2x /q jo^H? , Ix 2 -j- 6 
cta - a 2 -&\ + \ar + & 2 X^ + a 

Rationalize the denominator before differentiating. 



n V x 2 + 1 4- a dv / . 2 a? 2 + 1 \ 

V a 2 + l _ a? dx \ Va^Tl 



10. y= 1-ft 2 . <fr = -2tt(2-s») 

\(14-« 2 ) 3 ^ Vl -x 2 V(l+^) 5 



/— n dv / i n - 1 Vl — X 2 — # 

11. 2/ = (x+Vl-^ 2 ) M . ^ = 71(X + Vl-X 2 ) , ' 

dx VI — x 2 

12. ^wVTT^+VT^). ^y^lfi. J — V 

dx x\ Vi_W 

_ (smnx) m dy _mn{smnx) m ~ 1 aos(mx — nx) 

(cosmx) n dx~ (cosmx) n+l 

u y _ VH^+VT3^ j d y _ 2 / X| 1 \ 

VT+a?- Vl-a? *" A Vl-zV 

15. V = tan"'? + log .JEEi dy = 2a^_ 

a \« + a cfcc a; 4 — cr 



MISCELLANEOUS EXAMPLES. 



16. y = sec 



— eor> 1 , 



vV 



;r 





41 


dy_ 1 




da; Va 2 - 


-or 9 



17. /(a) = (a 2 -fa 2 ) tarr 1 -. f{x) = 2 a; tan" 1 - + a 



a 



§1= _ 


x sin 


l a 


dx 


Vi- 


a^ 


dy _ 


2 





18. y = Vl — a?i 



19. w = sin 1 - — -• 
* . 1 + a? da; 1 + a,- 2 

20. Find — (e ax sin 7 " ra) . 

da; 

/'(a?) = e ax sin m_1 ra5(a sin rx + mrcosra) . 

21. ?/ = log(2a;-l + 2Vaf'-a;- 1). ^ = * 

da; Vaf — a; — 1 

ao _i3 + 5cosa; , 4 da; 

22. ?/ = cos J — dy — 



5 -|- 3 cos a; 5 +3 cos a; 

23 . f(x) = e (a+x)2 sin x. f (x) = e (a+x)2 [2 (a + ®) sin a; + cos a;] . 

Tito*- 1 da; 



24. y = log [log (a -f bx n ) ] . dy 



(a + bx n ) log (a -f to") 



25. 2 / = log('^Y-Jtan- 1 a;. dy = ^ 4 - 

K l — x 1 — a; 4 



nn . l 3-\-2x , da; 

26. ?/ = sin- 1 — — dy — 

Vl3 Vl -Sx-x 2 

27. y = e x * tan" 1 x. ^ == e x * | * - + af tan" 1 as(l + log x) 
J dx L 1 + * - 



x-y 

28. a? = e~!T. 



. dy log a; 

log a;; ,\y = 



1 -{- log x da; ( 1 -f- log x) 2 



42 DIFFERENTIATION. 

X 2 



29. y 



1+-S 



1 + 



1 + etc. to infinity. 
.'.dy= ± 



1+2/ Vx 2 



30. y = \og(x+-Vx i -a 2 ) + sec- 1 -. ^ = lJ a?+q . 

a dsc cc \ # — a 

31 „ = log Vi - g 8 + ^V^ . gy = V^ 



32. y = logJ^^± 






V 1 +A3 2 — flJ 

33. Given/(*)=3x 2 -a + 6; to find /(y), /(a), /(2), /(0), 

/(^ + 2/)- 
When, in connection with the symbol /(a), expressions like 
f(y)* /(«) j /( 2 )> /(°)j f( x +.y) are employed, they denote 
respectively the results obtained by substituting y, a, 2, 0, and 
x + 2/ f° r * i n the value of f(x) . Thus here /(y) = 3y 2 — y + 6, 
/(a) = 3cr-a + 6, /(2) = 16, etc. 

34. Given (/>(>) = 4 a 3 -a 2 ; find ^(2), <^(6), <^(1), 0(3), 0(0). 
Given f{x +y)= <f» ; to find f(x) , /(y) , /(5) . 

35. Find d/(a + ic). /'(« + x)d(a + x), or f'(a + x)dx. 

36. Find df(ax). f'(ax)adx. 

38. Find df{x + y) . /'(a + y) (da + dy) . 



CHAPTER III. 
INTEGRATION. 

58. A function or variable is called the Integral of its differ- 
ential. Thus, x 3 is the integral of 3x?dx ; and f(x) -f- C, C being 
any constant, is the integral of f'(x)dx. 

Integration is the operation of finding the integral of a 
differential. 

The problem of differentiation and the inverse problem of 
integration ma}' be stated also as follows : 

That of Differentiation, or of the Differential Calculus, is, To 
find the ratio of the rates of change of a function and its variable. 

That of Integration, or of the Integral Calculus, is, Having 
given the ratio of the rates of change of a function and its vari- 
able, to find the f auction. 

The sign of integration is J . Thus, J in J ±x\lx indicates 

the operation of integrating A:x\lx. Hence, d and J , as signs of 

operation, neutralize each other. For example, J cl^x 3 ) =x*, and 

d J 3 xdx — 3 x^dx. The whole expression J 4:X*dx, read '-the 
integral of 4or 3 c?#," represents the integral of ±x?dx. 

59. Elementary Principles. 

I. Since clC=0< C being any constant, J 0=C. 

Hence, as may be added to any differential, the general form 
of its integral will contain an indeterminate constant term. 

In the Applications of the Calculus, this constant term is 
eliminated, or determined from the data of the problem. 

II. Since d(ay -\- ac) = ady, 

.-.j ady = j d{ay + ac) = a(y + c) 

= a)d(y+c) = ajdy. 



44 INTEGRATION. 

Hence, a constant factor can be moved from one side of the 
sign of integration to the other without changing the value of the 
integral. 

III. Since d(x — y -f z + c) = dx — dy -f- dz, 

.-.J (dx — dy-\-dz)=:x — y-\-z-\-c 

— J dx — J dy -+- J d(z + c) 

— j dx — jdy -{- j dz. 

Hence, the integral of a sum of terms is equal to the sum of 
the integrals of the terms. 

60. Fundamental Formulas. Since integration is the inverse 
of differentiation, general formulas for integration may be 
obtained b} r reversing the general formulas for differentiation. 

dx , ,„ dx 



1. ( — = loga? + logc,* v d(logx-j-\ogc) = — 

J X X 

2. faafcto = ^— -f <7, \'df^^ + C] = ax n dx. 
J n + 1 U+l J 



3. J a* logacte = a x + (7, •,• d(a* -f- C) = a x \ogadx. 
J e x dx = e x -j- (7. 

4. J cos x dx = sin x -\- C\ V d(sinx + C) = cosxdx. 

5 . J — sin xdx = cos x -f (7. 

6. J secrxdx = tana; + 0. 

7. J — cosec 2 #cfce = cotic -+- C. 

8. J seccctan#cfcc = seca;-f- C. 

* When the integral is a logarithm, it is customary to write the indeter- 
minate constant term as a logarithm. 



FUNDAMENTAL FORMULAS. 45 

9. J — coseco?coticc7ic = cosec# + C. 
*10. I s'mxclx= versa; -f (7, or — cosa; -f- C 
*11. I — cos x dx = covers x + (7, or — sina;-f- (7 f . 
12. f da; =sin- 1 a;+C. 

•/ Vi-.t 2 

*13. f_ ZJE_ = cos" 1 a;+ (7, or - sin- 1 a; + C. 

14. f-^-^taii-^+a 

J 1 + ar 

*15. fjZ^. = C ot ^ + C, or - tan" 1 ;*; + C. 

J 1 + XT 

1 6 . f da? =sec-^+G. 
J x^/x 2 — 1 

*17. I — ~ c,a? == cosec -1 ^ -f (7, or — sec -1 a; -f C". 
J a- Var 9 - 1 



18. f da; = vers- 1 x 4- C. 
J V2 x — x 2 

/— dx 
— ^ = covers -1 a; + (7, or — vers -1 a; -f- C ! . 

V2 x — x 2 

The differentials in these nineteen formulas are the funda- 
mental integrable forms, to one of which we endeavor to reduce 

* Two integrals having the same or equal differentials must change at 
the same rate ; hence they must be equal, or have a constant difference. 
The constant difference between the variable terms of the integrals in the 

last four starred formulas is evidently - ; for, when x < -, 

J 2 2 

cos -1 a: + sin -1 a:=r-, cot _1 .r + tan _1 a: = -, etc. 
2 2 

The starred formulas are not necessary, since the second integral in 
each is given by a previous formula. 



46 INTEGRATION. 

,every differential that is to be integrated. The processes of the 
Integral Calculus are largely a succession of transformations 
and devices to effect this reduction. 

61. To facilitate the application of formulas 1 and 2, the}* 
may be stated as follows : 

I. The integral of a fraction ivhose numerator is the differen- 
tial of its denominator is the Naperian logarithm of the denomi- 
nator, plus a constant. 

II. Whenever a differential can be resolved into three factors, 
— viz., a constant factor, a factor which is a variable with any 
constant exponent except — 1, and a factor which is the differential 
of the variable without its exponent, — its integral is the product 
of the constant factor into the variable ivith its exponent increased 
by 1, divided by the new exponent, plus a constant. 



Examples. 
Find 

1. (axHx. Ans. — -f C. 

7 

2. jbxhdx. ■§&#§ + C. 

3. $2xidx. f«l.+ C. 

4. J (J axi — | bx%) dx. axl — bxl -+- C. 

5. C^.= Cix- 5 dx. - x -* + a 



x 2 3x? 



7, 



f^r- 2V5 + C 

J V x 

2 

-yjx 



8. f/W+J-W ^__L + (7. 

J \ xi y 4 ^/ x 



EXAMPLES. 47 

a 
(1 — ri)x n ~ l 



q Cadx a ^ 

J x n 



10. f&(6«a: 2 + 8to 3 )§(2ax + 4to 2 )dx. 

Since d(6 ax 2 + 8 to 3 ) = (12 ax + 24 to 2 ) ax, we see that 
the differential factor (2ax + 4 bx 2 )dx must be multiplied 
by 6 to make it the differential of the variable 6 ax 2 + 8 to 3 . 

.'. r6(6ax 2 + 8 6or 5 )i(2ax + 4 6a; 2 )cZx 

= f-(6a<B 2 + 8&ar J )§(12aa; + 2±bx 2 )dx 

=—(6ax 2 + 8bx s )% + C. § 61, II. 

16 

11. I \_a(ax -\-bx 2 )%dx + 2b(ax-{-bx 2 )*xdx~\. 
I \a(ax-\- bx 2 )?dx + 2b(ax -\-bx 2 )*xdx~] 

= f(aa; + bx 2 y*{a + 2 to)ox == f (ax + to 2 )! + C. 

12. C(2a + 3bx) 3 dx. —(2a + 3to) 4 + 0. 

13. f f ^ , -|(a 2 + x 3 )* + (7. 
J (a 2 + ar 5 )i 3 V ' 

14. f (1 + fx)*dx. ^.(l . + |»)l+ 0. 

15. f(6x 4 +2x 2 -5)(3x 2 -l)ax. i^_iI^ + 5x-f-<7. 

. _. log(x — a) -f-logc, or log [(x — a)c]. 

x — a 

17 rx n ~ l dx 
' J a + fcx 71 ' 

/'x n ~ 1 dx 1 Cnbx n ~ l dx 1 , , , r n \ t i 
= — I = — log (a 4-to n )+logc. 
a + to n w6 J a + bx 11 nb 

18 -/ToSr5- log[(10*» + 15)*c]. 



48 
19 
20 



JGSfTEGliATION. 



•/; 



f. 



2bx 2 dx 
ae -j-foe 3 

bbxdx 



Sa-bbx 2 



log [(ae + &<*?)*<;]. 

W £ 

B (8a-6bx 2 )& 



01 f 5(2a-a 2 ) 3 d« „[" 2a 3 , 6a 2 , „ , x 2 ~j , n 

21 J s» ' o^-_ + - + 6alog,-^J + 0. 



23. J cotajdjc = | 



§& 3 ^ 



cos x dec" 
sin a? 

dx 
x 



2 4. f_*L = r 

J xlogx ^ «/ \ogx_ 
25. C^/2pxdx. 

26.f2^yft +1 J d y 



ax 

(3a^-^ 3 )^ 



& 2 icl + •&&»¥ - ^ggft + (7. 
log (c sin a). 

log (logo?) + logc. 
farv/2pa + 0. 



g(2/ 2 +P 2 )§+0. 



97 / *— (2 ax — x 2 )dx 



2* 



. C(2x i -3x 2 +l)h(x s -^x)dx. -^(2x A -3x 2 +l)l + C. 

29. I sin x cos x da?. 

30. f(logcc) 3 ^. 
J x 

31. j a 3x logadx[=-j- J aMogaScfa;] 
.J"(log»)' 
. I a 4 * da. 



32 



,dx 



ra + 1 



Jsin 2 aj + C« 
£(loga) 4 -f logc. 

ia^ + C. 
(loga^^ + logc. 



4 log a 



EXAMPLES. 



49 



34. e n dx. 



36. Cae bx dx. 

37. Cca^dx. 

38. I cos hnx)dx =— ( cos (race) 
J [_ m*/ 

39. I sec 2 (mcc)d£c. 

40. I sin 4 x cos x dx. 

41. i sin 3 (2a?)cos(2#)da3. 

42. Jcos 4 (3a?)sin(3aj)da;. 

43. I sec 2 (o? 3 )^ dx. 

44. I 7 sec 2 (a; 2 ) a; da?. 

45. I logx 



mda? 



dec 
as 



46. 



/sin a? da; _ 1 /*— b sin a; cte 

a 4- & cos a? b J a-\-bcosx 



+ 6 cos x 

47. I 5 sec(3a?) tan(3a?)da,\ 

48. I 5 cos (a •+- bx)dx. 

49 . J 4 cosec (aaj) cot (aa?) da;. 



we n + C. 



sec#-b C. 



'■e b * + C. 



C „2z 



2 log; a 



a 2x + C. 



— sin (ma?) + C. 
m 



— tan (ma?) + (7. 
m 

|sin 5 a?-{-C. 

|sin 4 (2a?) + <7. 

__L C os 5 (3a?)+0. 

^ tan x s + (7. 

-|tana? 2 -f- O. 

^(logaj) 2 + logc. 

log r 

(a -\-b cos a?) 5 

f sec (3 a?) + (7. 



5 

-sin(a + &a?) -f- (7. 
6 



cosec (ax) + 0. 



50 

50. Ce cosx sinxdx. 

51. I e 2s[nx cosxdx. 

/(I +cosx)dx 



INTEGRATION. 



D'2 



x 4- since 



53 



2 a* da* 



r xdx r i r 

J 1 4-a- 4 [_~~2J l + (^) 2 _ 
54. \(l+x){l —x 2 )xdx. 



±e 28inx +C. 

log [ (a; 4- sin aj ) c ] • 

£ tan" 1 x 2 4-0. 



/V»2 /v»* /yid /VH> 

y~y y~y 



a-V J 



56 



/ a; 4 da; 
a* 2 4-1 



57. p 



|! + a;4-21og(a--l)4-0. 
x 4- tan~* a* 4- O. 



a;' 1 x dx 



(a + &aJ*) : 



(a 4- bx n y 

rr> /^ bX Z dX 

° J 3 a* 4 4- 7* 



a. 



&n(l — m) 

log [c (3 a; 4 +7) A] 



62. Auxiliary Formulas. By integrating the equations in Ex- 
amples 1 and 2 of § 57, and those in Examples 27 and 28 of 
§ 39, we obtain the following auxiliary formulas for integration : 



(a) C-te = I t an- 1 -+C. 
J x l 4- a a a 

v ' J x 2 -a 2 2a s a*4-a B 

( C ) f dx = sin- 1 - 4-0. 
J Va 2 - x 2 a 

(d) f dx = log (a* 4- Va* 2 ± a 2 ) 4- log c. 

J Va,* 2 ± a 2 



AUXILIARY FORMULAS. 51 

(e) f — ^ = -sec- 1 -4-Q. 

J x^/x 2 — a 2a a 

(f) C dx ^yers- 1 - 
•^ V2 ax — x 2 a 



■ C. 



If the differentials in these formulas were negative, we would 

evidently have cos 1 -, or — sin -1 -, in place of sin" 1 - ; 
a a a 

log 37 "' -, or log , in place of log ~ ; etc. 

x — a a — x x+ a 



Examples. 

1. Deduce formulas (a), (c), (e), and (f) of § 62, from 
formulas 14, 12, 16, and 18 of § 60. 

dx 



(0 J^*--/ •=_ 

J V or — xr^ L xr 



= sin _1 --|- G. 

a 



dx 

a 



, . C dx 1 C a 1 i# 

(e) I =- I ^_sec- ] --f 

J x-\Zx? — a 2 aJ x x 2 . a a 

a \ cr 

2. Deduce formulas (b) and (d) of § 62 from formula 1 of 

§ 60. 

Since — -= — f 

x 2 — a~ 2 a \x — a x + a, 

/dx 1 C dx 1 r dx 1 , x—a.p 

x^ — a 2 2aJx—a 2aJ x + a 2 a a x-\-a 

To deduce (d) , assume Vx 2 ± a 2 = z — x. 

.'. ± a 2 = z 2 — 2xz. 

. / N 7 , dz dx dx 

. . (z — x)dz = zdx, or . — = 



x Vx 2 ± a 2 



Itan- 1 ^ + C. 
be c 



52 INTEGKATION. 

.-. f — ^— = f— = log(zc) = log[(a? W^^ 2 )c]. 
J Va? 2 ± a 2 J z 

3. Find C dx f=-f d(bx) 1- 
* ma Jc 2 + &v[_ &J C 2 + (&*) 2 J 

Here a? of formula (a) = 6a?, and a = c. 

4o r cto nr d(bx) i i sin -i^ +0 , 

J Va 2 c 2 -6VL 6J V(ac) 2 -(6a?) 2 J & ac 

5. f dx f_ f d ( cx "> 1 J-sec-^ + O. 

6 f — ^ r_ i r z^(^) 1 i 

' J V^^cV L °J -V8(cx)-(cx) 2 J c 



covers -1 \- C. 

4 



, r dx r r d(Vix 

J V7a? 4 — bx' 2 \_ J s/lx^Tx 1 
;. f *! "= f. 



i„ -.(4D +0 . 



>S 



-: X + 



IV 



/ 



da? 



-yJ{CLX — X 2 ) 



f-m 



dx 



Vf-f-I 



10 



r dx r r dx 

' J a? 2 + 4a? + 5[_~J (a? + 2) 2 + l 



11 



12 



13 



■ji 



2 a? 2 — 2^ + 1 



■/; 



dx 



x-Vtfx 2 - a 2 



■s 



dx 



^(2abx-bV) 



V5 



sm * \- G. 



tfm~\x + 2) + C. 
tan- 1 (2a?-l)+0. 



1 ,bx . n 

-sec 1 f- C. 

a a 



1 -\bx . -,* 

-vers l h G. 

6 a 



EXAMPLES. 



J a/83 



15 



16 



V8 -4ar J 
dx 



Jd 



f a4 dx 



1 

3^ V2 - (a;§) : 



isin-^| 



53 



V3 x - x 2 - 2 



f-R 



VF 



3\ 2 



siii- 1 (2a;-^3)+0. 



/: 



dx 



Vl+a^ + aJ 



-/- 



cte 



vc-+ar+f 



log [a; 4 - 4 V^ 2 + a?+ 1 ) + logc. 



ar 3 — 1 
i l og-——-\-logc. 
x 6 4 1 



18 



19 



20 



2a : 



tan- 1 -, + 0. 



2 . _!2oj + 1 

- — tan l ■ — 



V3 



V3 



a 



r^dx 

J x 6 -l 

/' xdx 
tf + x 4 ' 

r dx 

' J 1 + x + x 2 ' 

/ —dx 
yfn (c 2 — x 2 ) 

21. f___^L_. 
•* ^Jax — x 2 

n ~ /^ —xdx / c -i2x.fi 

22. | - • wcx — x- vers * f- C. 

(*_^xdx_ ._ r c — 2x — c , _ r (c-2a;)(?a; c f * da; 
J ^/cx — x 2 J 2 -\/cx — x 2 J 2^/cx—x 2 2 J ^/cx—x 2 



n~i cos -1 -4 (7. 

c 



vers - \-C- 

a 



23 



24 



/: 



dx 



s 



■Vox 2 — b 
dx 



— -\og(xVa +-\/ax 2 — b) 4 logc. 
-\/a 



V 2 ax 4 a^ 



log (x-\- a 4 V2 aa; 4 x 2 ) 4 logc. 



54 



IMEGKATION. 



25 



■Si 



dx 



x 2 -{-Ax 

an / X (XX 

27 f 3 da; 

* J 4+9x 2 

"J a 2 + 



log hiogc. 



a 2 



2 



Jtan-^ + C. 



29 



f 3ar*c 



ar 



da; 



VV-9; 



30. r — ^_ 



31 



da? 



^tan- 1 - + -log (a 2 + a? 2 ) + C. 

a a '2 



Jvers- 1 (18a^) + 0. 



a 



/ " 5 da : 

"^ «V3a^ 



82 



r(a^-a 2 )^ 



da?. 



VSsec-^ajJ-V+O. 



(a* -a 2 )* -a sec" 1 - + C. 



r(a^-a 2 )^ a , == r (a^-ft 2 )da; = r a:da; r a 2 da? 

•J a* *^ x-y/x?—a 2 J -y/of—a 2 ^ xs/x^—a 2 



A fractional differential may often be separated into integra- 
te parts, or reduced to an integrable form, by multiplying its 
numerator and denominator by the same quantity. 



J Vi - x L J Vi 



/: 



a?) dx 



sin- 1 a;-(l-a3 2 )§+ C. 



-Vx-j-a 



34. I y M ^^ dx. 

x~\/x — a 



sec -1 - -f log (x + Va? 2 — a 2 ) + log c. 



0f . C dx J~_ ^_J^da?_ 
J ' J (1 - aj")i \_~J (x~ 2 - l)lj 



Vl-a? 2 



<7. 



TRIGONOMETRIC DIFFERENTIALS. 



55 



63. Trigonometric Differentials. The following trigonometric 
differentials are readily reduced to known forms. The forms 
in the first seven or eight examples should be especially noted. 



l.Find 



r dx 
J sin q 



Ans. log tan (%%) -\- C. 



/ dx _ r dx C sec 2 {\x)^dx 
since J 2 sin (^x) cos (^ x) J tan (-|- as) 
= log tan {\x)-\- C. 

/ dx ~ _ r dx 
cos x J 



sinf - -f-as 



U+5 



log tan I - 4- - ) -f C. 



o C dx T_ f* sec 2 as c?as ~l 
J since cos iu |_ J tana? 

4. I cos 2 asdas[ = I (-J- -J- -J cos 2 as) das] 

5. I sin 2 as c?as. 

6. fcotx^r=f cosa!da; "|. 
c/ J sin a; 

7. I tan as das. 



dx 



dx 



,2^«^o2, 



suras cos" as 
9. I sin 5 as das. 



log tan a; + (7. 
--hi sin 2 as -f (7. 
|-i.sin2a?+0. 

log sin tc + (7. 

— log cos cc 4- (7, or log sec as 4- C. 

tan as — cot x -\- C. 

/(sin 2 as 4- cos 2 as) das (*, „ . „ x 7 

■* rV * — = I ( see-as +cosec 2 as) das. 
snr as cos- as J 



I sin 5 as dx = I (1 — cos 2 as) 2 sin as dx 

= — I (1 — cos 2 as) 2 d( cos as) 



— cos as + -| cos 3 as 



* cos as 



+ 0. 



56 INTEGRATION. 

In like manner, j sin n xdx and j cos n xdx can be found, when 
n is an odd positive integer. 

•/< 

. I Qos z xdx. 

. I sin 5 a; 

j sin 5 x cos 3 a; da; = J sin 5 a;(l — sin 2 x)eosxdx. 

In like manner, j sin m xcos n xdx can be found, when either 
m or n is an odd positive integer. 

14. j sin 3 a? cos 3 a; die. isin 4 a? — -|-sin 6 £c-|- C. 

15. I cos 4 a;sin 3 a?cfa;. — ieos 5 a;+-i-cos 7 a; -+- (7. 



10. i swPxdx. 
11 

12. | cos 5 xdx. 

13. I sin 5 a? cos 3 a; da;. 



J COS 3 a; — COSfl? + C. 

sin x — isin 3 #+ 0. 

sinx — -|sin 3 cc4--|-sin 5 fl?+ C. 

Jsin 6 a? — ^-sin 8 ic -f- C. 



16 



17 



10 



/cos 3 a; cZa; 
sin 4 a? 

•/ COS 2 £C 

/sin 3 a; 
cos 8 a? 

/sin 3 a; 
cos 7 a; 

/sin 2 x 
cos 6 a? 



(1 — sin 2 o?)d (sin x) 
sin 4 a? 



1 



+ C. 



sin x 3 sur a; 
seca;-f-cosa; + C. 



dx. 



^sec 7 x — |-sec 5 ic+ C. 
i sec 6 a; — ^sec 4 a;+ C. 
±-tan 5 a; + -J- tan 3 a; + C 

/'- — -^-dx= I tan 2 a; sec 4 a? da; = ( tan 2 a; (tan 2 a; -f 1) sec 2 a; da;. 
cos 6 a? J J 



20. | ^^dx 

cos 1 



EXAMPLES. 57 

In like manner, dx or — dx may be integrated, when 

cos" a? sm n x 

m — n is even and negative. 

21. C^^dx. _i co t 3 a;+C. 
J sin 4 a; 

22. f^Hlfcfa.. itan 6 aj + C. 
J cos 7 a? 

23. f-^-. tana; + 1 tan 3 a; + 0. 
J cos 4 a; 

24. J tan 2 a; daj[= | (sec 2 a;— 1) da?], tana? — x + C. 

25. I cot 2 a?da?. — cota? — x-\-C. 

26. I tan 3 a?da?. ^tan 2 a?-|-logcosa?-|- O. 

27. J cot 3 xdx. — i cot 2 x — log sin a; + O. 

28. J tan 5 a; da;. 

I tan 5 xdx= J (sec 2 a; — l)tan 3 a?da? = J tan 4 a? — J tan 3 xdx 
= itan 4 a? — I (sec 2 x — 1) tan xdx 
= itan 4 a? — -J- tan 2 a; — log cos x-\- O. 

In like manner, tsiu m xdx and cot w a?c?a? may be integrated, 
when m is a whole number. 

29. I tan 4 a?da?. |-tan 3 a? — tana? -\-x -f- C. 

30. I cot 4 a?da?. — icot 3 a? + cota? -f- a? -f 0. 

31. I tan 6 a; da?. J-tan 5 a? — itan 3 a?-f- tana; — x + C. 



58 INTEGRATION. 

32 . J cot 5 x dx. — \ cot 4 x 4- \ cot 2 x -f- log sin x+C. 

64. Definite Integrals. All the integrals yet found contain the 
indeterminate constant term (7, and are called indefinite integrals. 

When C is eliminated, or determined for any hypothesis, the 
integral is called a definite integral. 

When, from the data of a problem, we know the value of the 
integral for some particular value of its variable, C can be 
determined. For example, suppose that du = 2axdx, and that 
u = when x= 2. 

Since du — 2 ax dx, u = ax 2 -f C. 

Since u = when x = 2, = 4 a + (7. 

Hence, C = — 4 a, and m = ao 2 — 4 a, a definite integral. 

If, in any indefinite integral, two different values of the vari- 
able be substituted, and the one result subtracted from the 
other, C is eliminated, and the integral is said to be taken be- 
tween limits. The symbol for the definite integral of <j>(x)dx 

between the limits a and b is f cf>(x)dx; a and b are called the 

limits of integration, a being the inferior and b the superior 

limit. The symbol I indicates, that the following differential 

is to be integrated ; that a and b are separately to be substituted 
for the variable in the indefinite integral ; and that the first of 
these results is to be subtracted from the second. 

In what precedes, we assume that the integral is continuous 
between the limits a and b. 

In the indefinite integral, neither limit of integration is fixed 
upon. In the first form of the definite integral, only the inferior 
limit is determined, and the integral is still a function of the 
variable. In the definite integral between limits, both limits are 
fixed, and the integral ceases to be a function of the variable. 

Examples. 

1. Given dy = (1 + %ax)?dx ; find the definite integral on the 
hypothesis that y = when x = 0. 



EXAMPLES. 



59 



Here, - i(l + *«)• + </; , .0 = ^+0. 



* 27a V 4 ; 27a 

2. Given dy = (x 3 — b 2 x)dx-, find the definite integral, if 
y = when x = 2. 

Ans. 2/ = ^l-^ + 26 2 -4. 
^4 2 

3. Given ay = ~ — ; find the definite integral, if y = 



a; 2 — cc 



when a?== 1, 
4. 



^Ins. 2/ = log (2ce — x 2 ) 



:. Find f 



nxdx. 



Ans. -(b 2 -a 2 ). 



nxdx= -x 2 -f- (7; 



irf+0 






a^+O 



= -6 2 +(7. 



D 

.-.£nxdx = 'p 2 + C-(^a 2 +C^ = 7 j(b 2 -a 2 ). 

5. (\x 3 dx. 

6. I (asr 2 — or 3 ) aa;. 

7 p da; 

Jo a 2 -+- cc 2 

8 p dx 

Jo Va 2 - a 2 
sin (9 0*0 



. I — = I (cosO) -smOdO 

Jo COS 2 (9 \_ Jo V 



24. 
12' 

7T 

la' 

7T 
2' 

V2-1. 



B 



-x 2 + C I denotes the value of -x 2 + C when x = a. 
2 L 2 



60 



INTEGRATION. 



10. x n dx. 



e~ ax dx. 



12 






2r 2-V2rdy 



5»+i _ a n+1 

71 + 1 

1 

a 

Br. 

256 tt6 9 
315 a 4 ' 



V2r — y 

13- f-^(3/ 2 -6 2 ) 4 ^. 

c/ -6 a 4 

'Jo 1-f # 4 * 

sin 3 # cos 3 # da;. 

Applications to Geometry and Mechanics. 
65. Rectification of Curves. From § 16, Cor. 2, we have 



7T 

4* 

J_ 
12* 



ds = -ydx 2 + (fa/ 2 ; 



This equation is a general formula for the rectification of any 
plane curve ; that is, for finding its length. 

Examples. 
1 . Rectify the semi-cubical parabola y 2 = asc 3 . 
Tj dy _ 3 ax 2 . . d?/ 2 _ 9 aa; 
616 dx~~2y~' ' 'dx>~ 4 



,'. s = jYl + ^j£Y(te = J f(4 + 9 ax)hdx 



(4 + 9qg)l . e 

27a 



(1) 



So long as the point from which s is measured is undeter- 
mined, C must be indeterminate. If the length of the curve 
be estimated from the origin, s = when x = 0. 



AREAS OF PLANE CURVES. 61 

Substituting these values of s and x in (1), we have 

27a 27a K } 

If, in (2) , a = 1 and x = ±,s= 2^ ; that is, the arc of y 2 = x 3 
that lies between the origin and a; = f , is 2 Y 2 T in length. 

For the length of the arc, the abscissas of whose extremities 
are b and c, we have 

V& y 27a 

2. Find the length of a branch of the cycloid 

x = r vers -1 -^ — V2 ?*?/ — 2/ 2 « 
r 

dy 2r-y \ dy-J 

,.2 A K(Fi g .14)=2p(l + ^ 

= 2V2r f (2r - ^)"5% = Sa- 
lience the length of a branch is eight times the radius of the 
generating circle. 

66. Areas of Plane Curves. From § 14, we have 

dz = ydx ; 

.-.2= \ydx = \f(x)dx. 

This equation is a general formula for finding the area included 
between an} r plane curve and the axis of x. 

In applying this formula, it must be borne in mind that, area 
above the axis of x being positive, area below it is negative. 

For the area between a curve and the axis of y, we evidently 
have 



*% 



=J xdy. 



62 



INTEG11ATION. 



Examples. 

1 . Find the area between y 2 = 2px and the axis of x. 
Here z = J ydx = J (2p#)2djc = f#V2px -f- C—\xy + 0. 

If the area be reckoned from the origin, z = 
when a; = ; 

.-.(7=0, and 2 = %xy. 

Hence the area oap = | oapr; and the area of 
the segment nop is two-thirds that of the par- 
allelogram MNPR. 



E 




I 


H 



Fig. 15. 



If oh = a, and oa = b, 



area bnpd 



2 C\/2pxdx = I V2p (61 



at). 



2. Find the area of y -■= x 3 -f- ax 2 between the limits x = — a 
and x = ; also between the limits x = and cc = a. 

J.71S. yV^ 4 ; T2« 4 . 

3. Find the area of the hyperbola xy = 1 between the limits 
x = 1 and x = a. 

Area = log a ; that is, the area is the Naperian logarithm of 
the superior limit. It is because of this property that Naperian 
logarithms are sometimes called hyperbolic logarithms. 

4. Find the area inclosed by the axis of x and the curve 
y = x — x 3 . 

The inclosed area lies below the axis of x, between x = — 1 
and x = 0, and above it, between x = and x=l. These two 
portions being numerical!}- equal, the result obtained by inte- 
grating between x = — l and x=l is 0. To find the required 
area, obtain the area of each portion separately, and take their 
numerical sum. 

Ans. 1. 



AREAS OF SURFACES OF REVOLUTION. (53 

5. Find the area of the ellipse a?y 2 -j- b 2 xr = a 2 b 2 . 



Area = -4 J Ver — x 2 dx. 
a Jo 



4 I Va 2 — x 2 dx = ira 2 ; 



for it evidently equals the area of the circle whose radius is a. 



area = -ira 2 = irab. 
a 



6. Find the area intercepted between y 2 = 2px and x 2 = 2py. 

Area = | '^J'lpxdx — j — dx — -^> 
Jo Jo 2p 3 



7. Required the area intercepted between y = — and 

J. -T- XT 
_ X 

y ~ X Ans. log 4 — -f . 

67. Since z = j ydx= J f(x)dx (§ 66) , the integral of f{x)dx 
can be represented graphically by the area between the curve 
y—f{x) and the axis of x. Hence, when ) f(x)dx cannot be 

Xb J 

f(x)dx can be determined approximately by com- 
puting geometrically the area of the figure formed by the axis 
of a?, y =f(x) , x = a, and x = b. 



68. Areas of Surfaces of Revolution, 

Let s represent the length of the curve opn, y 
and S the surface generated by its revolution 
about ox as an axis. 

To obtain a general formula for the value of 
#, let As = arc pp' ; then AS = the surface 
traced by As. 

limit f As 
As = ° chord pp' 




Since 



= i; 



§ 4b. 



64 



1NTEGKATION. 



. limit [" surface traced by As 
As = 0[^ sur f ace traced b}^ pp' 



, or 



, limit 

AsiO 



As 



surface pp' 



As 



,=i; 



limit 



it [AS1 



limit 
As = 



2 ) As 



.-. — = 2tt2/. 
as 



.-. S = 2 7T Cyds = 2 iz Cyfl + 



dot? 



dx. 



When the axis of revolution is the axis of y, we have, simi- 
larly, 



8 



2-rr I xds = 2 7r J x(l-\ 2 }dy. 



Examples. 

1 . Find the surface of the sphere. 
Here the generating curve is x 2 -\-y 2 = r 2 . 

■■■ s = 2 €K l+d £) idx=2 €i i+ f) idx 

= 2 it J rdx = 4irr 2 . 

2. Find the surface of the paraboloid ; that is, of the surface 
traced by the revolution of a parabola about its axis. 

S = 2 Tf £y (dx 2 + dtf) 1 = 2 ttJj (l + |^*dy 

= 2tt 



dtf 
[(& 2 +i> 2 )^-p 3 ]. 



VOLUMES OF SOLIDS OF REVOLUTION. 



65 



69. Volumes of Solids of Revolution. 

Let V represent the volume generated by the revolution of 
opw about ox as an axis. To deduce a general formula for the 
value of F, let Ax = ab ; then Ay = dp', and 
A V = the volume generated by the revolu- 
tion Of ABp'p. 

Now, volume abdp < A V< volume abp'm ; 

or Try 2 Ax < AV< ir(y-\- Ay) 2 Ax; 

AV 



.-. iry* < -— < TV (y + Ay) 2 ; 

Ax 



Fig. 17. 



dV 
dx 



Try 2 , or V— 7T I y 2 dx. 



(i) 



Or, to obtain (1), conceive the solid as generated by a varia- 
ble circle, whose centre moves along the axis of the solid, and 
whose radius is equal to the ordinate of the generating curve. 
With this conception, it is evident that, if dx = AB, dV=the 
cylinder whose altitude is ab, and the radius of whose base is ap. 



.-. dV= 7ry 2 dx, or V= -n- | y 2 dx. 



When the axis of revolution is the axis of?/, we have, simi- 
V— TT J x 2 dy. 

Examples. 



1. Find the volume of the prolate spheroid; that is, of the 
solid generated by the ellipse revolving about its major axis. 



c a r a b 2 

Here V= tt I y 2 dx = ?r I — (a 2 — x 2 )dx 



f(2«7T& 2 ). 



Hence the volume of the prolate spheroid is two-thirds the 
volume of its circumscribed cylinder of revolution. 



w, 



If a = b, V 
which gives the volume of the sphere whose radius is a. 



66 INTEGRATION. 

2. Find the volume generated by the revolution of y 5 = ex 
about the axis of x, volume being measured from the origin. 

Ans. V= \^y 2 x = -| the circumscribed cylinder. 

3. Find the volume of the oblate spheroid; that is, of the 
solid generated by the revolution of the ellipse about its minor 
axis. 

Ans. V= ^7ra?b = % the circumscribed cylinder of revolution. 

4. Find the volume of the paraboloid; that is, of the solid 
generated by the revolution of the parabola about its axis. 

Ans. V= \-rtxy 2 , or \ the circumscribed cylinder. 

5. Find the inclosed volume of the solid generated by the 
revolution of y 2 — b 2 = ax? about the axis of y. 

Ans. V =^f\f- b ycly = ?§^. 
crJ-b 3 lo a 4 

70. From v = — and a = — (§ 33), we have the following 

dt dt v J & 

Fundamental Formulas of Mechanics. 

I. v== — ; .'.s = I vdt, and t= I — • 

dt J J v 

II. a =—;.*. v == I adt, and t = ( — 
dt J J a 

Examples. 

1. The acceleration of a moviug body is constant; find the 
velocity and the distance. 



V=fa 



dt = at+C=at + V , (1) 



in which v represents the initial velocity ; that is, the value of 
v when £=0. 

8= | vdt= I (at -\- v )dt = ± at 2 -\- v t -\- S Q , (2) 

in which s represents the initial distance ; that is, the value of 
s when t = 0. 



APPLICATIONS TO MECHANICS. 



67 



If the motion begins when t = 0, v = and s = ; hence (1) 
and (2) become 

v = at and s = ±-at 2 ; 



.' .t =\i — and v=V2as. 

These four formulas are the fundamental formulas for uni- 
formly accelerated motion. 

The acceleration caused by gravity is 32.17+ ft. a second, 
and is denoted by g. If we substitute g for a in the four for- 
mulas given above, we obtain the formulas for the free fall of 
bodies in vacuo near the earth's surface. 



2. By a principle of Mechanics, if ab be a vertical line, the 
acceleration of a body sliding without friction a 
along the inclined plane ac is #cos<£, in which 
$ = angle bac. Let s', v\ and t' represent respec- 
tively the distance ac, the velocity acquired along 
ac, and the time of descent ; then, from the for- 
mulas, 

v = V2as, and t =\j — , 

\ a 

we have 

i | •> s i 

v'.= w2qs' cos<f>, and t f =\ — 

J \gcos<f> 

Let s represent the vertical distance ab ; then 



s' = 



an 



COS(/> 

\ a cos- 6 cos 6 \ 



v' = ^2 gs 1 cos cf> = V2#s, 



2 s. 



gcos~cf> coscf> \ g 




Fig. 18. 



Hence the velocity acquired by a body sliding without fric- 
tion down ac equals that acquired by a body falling verti- 
cally down ab ; and the time of descent along ac is the 



time of descent along ab. 



COS(£ 



68 



INTEGRATION. 



3. Let ac (= s) be the vertical diameter of any vertical circle 
A abc ; then the time of descent from a along 

any chord ab (= s') is 




2 s' 



\#cos<£ 
since cos $ 



(Ex. 2), which equals 



& 



Hence the time of descent from a along any 
chord of the circle is the same as that along the vertical diameter. 

4. The acceleration varies directly as the time from a state 

dv 



of rest of the body ; that is, 
end of time t. 



dt 



a = ct\ find v and s at the 



Ans. v — ^cf; s = ^cf. 




5. When the velocity is a given function of the time, the 
time, velocit}', distance, and acceleration can be represented 

geometrically, as follows : 

Construct the locus of v =f(t), t being- 
represented by abscissas, and v by ordinates, 
the unit of t being represented by the same 
unit of length as the unit of v. Then, bj T § 66, 
the area between the curve and the axis of ab- 
scissas equals i vdt ; but, by § 70, s == I vdt. 

Hence, if abscissas represent time, and ordmates velocities, 
the area between the curve y =■/(*) and the axis of abscissas 
represents the distance traversed by the moving body. 

Again, if ph is a tangent at p, and ab represents the unit of 
time, dh represents the acceleration at the end of the second 
unit of time ; for it represents what would be the increase of 
the velocit} 7 , or ordinate, in a unit of time, if this increase be- 
came uniform at the end of the second unit of time. 

6. A body starts from o (Fig. 21) ; its velocity in the direc- 
tion of oy is constant, and in the direction of ox is gt ; what is 
its path? 



APPLICATIONS TO MECHANICS. 



69 



Let ox and oy be the axes of x and y, respectively ; 
then ^Z- 



dt 



, dx 
and — = 

dt 



Hence, y = ct, and x = ^-t 2 ; 

£4 

• 2c 2 
.' .11- = — x. 

y 9 




O) 



Fig. 21. 



Since (1) is the equation of a parabola 
referred to a diameter and a tangent at its 
vertex, the path of the body is an arc of a 
parabola. Hence, if it were not for the resistance of the atmos- 
phere, the path of a projectile, as a ball from a rifle, would be 
an arc of a parabola ; for its velocity would be gt along the 
action-line of gravhVy, and constant along the line of projection. 



7. The velocity of a body in the direction of ox is 12 t, and 
in the direction of oy is 4£ 2 — 9 ; find the ve- 
locity along its path onm, the accelerations 
and distances in the direction of each axis 
and along the line of its path, and the equa- 
tion of its path. 

Let v x1 v 9 , v s and a a , 04,, a s represent respec- 
tively the velocities and accelerations in 
the directions of the axes of x and y and 
along the path, whose length we will repre- 
sent by s. 




Fig. 22. 



Then 



^=v,= 12*, and %l = v=U 2 

dt dt 



9. 



dt \dt 2 dt' 2 v ; 



Vl6* 4 + 72* 2 + 81 = 4* 2 + 9. 



The accelerations are 



dv x 

dt 



dv 



12 : a v = ~^- = 8t; and a s 
y dt 



dv s 
dt 



St. 



70 INTEGRATION. 

The distances are 

x = Cl2tdt=6P, 

y= C(tt 2 -9)dt = ±t 3 -9t, 

and s= C(±t 2 + 9)dt = ±t s + 9t. 



(1) 
(2) 



Eliminating t between (1) and (2), we obtain for the equa- 
tion of the path, 



y=(i 



K - 9 >J- 



The form of the path is shown in Fig. 22. (See Weisbach's 
Mechanics of Engineering, page 148.) 



CHAPTER IV. 
SUCCESSIVE DIFFEBE1STIATI0N. 

71. Successive Derivatives. Since f'(x), the derivative of 
f(x) , is in general a function of x, it can be differentiated. The 
derivative of /' (a?) is called the second derivative of the original 
function /(a?) , and is denoted by f"(x). The derivative of 
f"(x) is called the third derivative of f(x) , and is represented 
by f"'(x) ; and so on. f n (x) represents the nth derivative of 
/(#), or the derivative of/ n-1 (a;). 

Thus, if 

f(x)=x*; f>(x)=£(x*) = 4x*; 

f"(x) = — (4^) =120?; f'"(x) =— (12a?) = 24aj; 

/"(»)= 24; / v (*0=0. 

/'(a?), /"(a), f"'(x), etc., are the successive derivatives 
of /(*). ' 

72. Signification of f"(x). Since /"(a;) is obtained from 
/""^(a;) in the same way that/' (a?) is from /(#), £7te ntfi deriva- 
tive of a function expresses the ratio of the rate of change of its 
(n — l)th derivative to that of its variable; and the (n — l)th 
derivative is an increasing or a decreasing function, according as 
the nth derivative is positive or negative. 

Cor. If a is finite, and /(«)*= oo, /'(<x)=cc, /"(a)=oo, etc. 

For, when /(a) = oo , f(a + h) is not go, however small h be 

taken. Hence, while x changes a very small amount from a, 

*/(a) represents the value of f(x) for x = a. The equation /(a) = oo 
means that f(x) increases without limit, as x approaches a as its limit. 



72 SUCCESSIVE DIFFERENTIATION. 

f(x) changes an infinite amount. Therefore, when x = a, f(x) 
must change infinitely faster than x does; hence f'(a) = 00. 
For like reason, iff (a) = 00, f"(a) = 00, etc. 

Examples. 

1 . Find the successive derivatives of x % -f 2 x 2 -f- x -f- 7. 
Let f{x) = x?+2x 2 -j-x + 7 ; 

then /'(a?) = — (ar 3 + 2a; 2 -f x + 7) = 3a,' 2 + 4a? + 1 • 
dec 

/"(a) = £ (3a; 2 + Ax + 1) = 6 s + 4 ; 

f'"(x)=£(6x + ±) = 6; 

and / IV (a;)=0. 

2. Find the successive derivatives of cx % -f- aa? 2 -j- a. 

3 . If f(x) = x 3 log as, prove that f iy (x) = — 

4. If f(x) == e ax , prove that /"(a?) = a' l e aa: . 

/ n (a?) is written out in accordance with the law discovered 
by inspecting f"(x) and /'"(a;). 

5. If f(x) = sin ma?, prove that f iy (x) = m 4 sin mx. 

6. If /(a?) = x* log cb, prove that/ VI (a?) = -^r- 

7. If/(aj)=af, prove that/" (aj) = af(l + log a?) 2 + af- 1 . 

8. If /(a;) = tan x, prove that /'" (x) = 6 sec 4 a; — 4 sec 2 a?. 

9 . If f(x) = log (e x + e~") , prove that /'"(a?) = - 8 ■ e 



{f + e-'Y 



10. If/(a?)= -^-, prove that/ IV (a;): 



1 — x (1 — a?)' 



[w, read " factorial n," stands for 1 X 2 X 3 X 4 X - X n, 



SUCCESSIVE DIFFERENTIALS. 73 

11. If f(x) = a 2 , prove that f n (x) = (log a) n a x . 

(-l) n - l \n-l 



12. lff(x) = log(l+ x) , prove that f n (x) = 



(1+*)' 



(-1) 3 |3 x (_l)»-iU_l 

13. If f(x) = ( 1 + x) m , prove that 

/»(#) = m(m -1) ••• (m - n +1) (l+x) m ~ n . 

73. Successive Differentials. The clifFerential of the first dif- 
ferential of a function is called the second differential of the 
function. The differential of the second differential is called 
the third differential. In like manner, we have the fourth, fifth, 
and nth differentials. d(dy) is written d 2 y, and read "second 
differential of y" ; d(d 2 y) is written c%, and read "third dif- 
ferential of y " ; and so on. 

dy, d 2 2/, cZ 3 ?/, etc., are the successive differentials of 2/. 

In differentiating 2/ =f(x) successively, dx is usually regarded 
as constant ; that is, as having the same value for all values of 
x. This greatly simplifies the second and higher differentials, 
and also the relations between the successive differentials and 
derivatives, and is allowable ; for, when independent, x may be 
regarded as changing uniformly. 

74. Relations between the Successive Differentials and 
Derivatives. 

H ,=/(*), !=/'(*); |(f) =/"(*); etc. 

dv 
If dx is variable, -2 is a fraction with a variable numerator 
dx 

and denominator, and we have 

fiu„\ _ d (dy\ __ dxdhj-dyd 2 x n , 

7 W ~dx[dxJ- ^ W 



74 SUCCESSIVE DIFFERENTIATION. 

But, if dx is constant, -^ is the product of the constant — 
dx dx 

and the variable dy, and we have 

W-s(*)-.S- w 

For the same hypothesis we have 

dx\da? J dx 3 dx* dx n 

Hence d n y =f n (x)dx n ; whence f n (x) is often called the nth 
differential coefficient 6ff(x) . 

For the rrypothesis that dx is constant, d 2 x= 0, and equation 
(1) becomes (2), as it evidently should. 

Examples. 

1. Find the successive differentials of x 4 . 
Let y = x* ; then dy = 4 x?dx. 

Differentiating this last equation, regarding dx as constant, 

we have 

d 2 y — (4:dx)d(x 5 ) = 12a? 2 eta 2 ; 

.-. d 3 y = {\2dx 2 )d{x 2 ) = 24xdx s ; 
.-. cZ 4 ?/ = 24cto 4 ; .\d 5 y = 0. 

2. Find the successive differentials of 5 or 3 -f 2 # 2 — 3 x. 

Ify=5x 3 +2x 2 -3x, d 3 y=30dx 3 . 

3. If y = sin a;, prove that d*y = sin a? dx 4 . 

4. If 2/ = log (ao?) , prove that cZ 4 ?/ = ~dx\ 

x 

5. If y = 2 a -y/x, prove that — % = — -• 

dx 3 4xi 

n Tij i 4.14- d s V 2 cos a; 

6. If v = log since, prove that — £■ = — • 

dx 3 snrx 



EXAMPLES. 75 



7. If y = x 2 log a;, prove that —4 = — 

dar x 

d*y 

8. If y = cos mx, prove that — ^ = m^cosma;. 

da; 4 

9. If y 2 = 2px, prove that — \ = -±- = — — • 

9 2 * dx 3 y 5 (2a;)§ 

dy 

dy_p. .&y_ = dx = P 2 . 

dx y' " dx 2 y 2 y* ' 

32/V ^ 
• c ii. = dx ^sp 3 ^ 3ps 

"dar 3 ~ y 6 f ~ {2x)i 

10. If ay + 6 2 ar 9 = a 2 6 2 , prove that ^ 7 = - -%- . 

dar a-^ 8 

11. If ar 2 + ?/ 2 = r 2 , prove that — =| = — — • 

dec 2 2/ 3 

12. If ?/ 2 = sec 2 x, prove that y + — ^ = 3 ^z 5 . 

dar 

13. If ?/ = e* sin a;, prove that ^ — 2^ + 2y == 0. 

dar da; 

14. Given s = 4Z 3 ; find i> the velocity, and a the acceleration. 

«. - ds dv d fds\ d 2 s 

Since v = — , a = — = — — = — -• 
dt dt dt\dt) dt 2 

Hence v = — — 12 1 2 , and a = — - = 24 £. 
dt dt 2 

15. If s = ct 2 + bt, what is the velocity and acceleration? 

d 2 s 
Ans. v =2ct -\-b ; a = — = 2 c. 
eft 2 

16. If ?/ 2 = 2pa;, prove that — - = — 

dy- p 

Here x is regarded as the function, and dy is constant. 



CHAPTER V. 

SUCCESSIVE INTEGRATION, AND APPLICATIONS. 

75. The general formulas for integration enable us to obtain 
the original function from which a second, third, fourth, or nth 
differential has been derived. 

For example, let d 3 y = bbdx 2, ; then 

d {S)- 5bdx - « 

Integrating (1), we have 

*2L = 5 bx + Gi ; .'\ d(^) = bbxdx + C Y dx : 
dx- \dxj 

dx 
.-. y = %bx* + i C,x 2 + a 2 aj + O s . 



Examples. 

1. Given dhj = 0, to find 2/. 

r.df^^C.dx; .-M=C 1 x+-G 2 \ 
\dxj dx 

.•.y = %C 1 x 2 +ax + C 3 . 

2. Given d 4 y = since dx 4 , to find ?/. 

cq J } = since da; ; .*. —*(-== — cosx + Ci ; etc. 
\dar/ dor 

^.ns. y = sin a? + \ C^ -f- -J C^ar 2 .-}- C 3 x -j- C t 



PROBLEMS IN MECHANICS. 77 



3. Given — ^ = 32? — x 3 , to find y. 
dx s 



V = to-^ 5 — i lo S x + £ Ci# 2 + C 2 a; + C 3 . 



4. Given — ^ = sin a;, to find y. 

da; 3 * 

y — cosa; + 1(7^ + C 2 x + 3 . 

5. Given d 3 y = 2ar 3 dar ? , to find y. 

y = logo; + i C^ar 2 + C 2 a; + C 3 . 

76. Problems in Mechanics. 

1. If the acceleration of a body moving toward a centre of 
force varies directly as its distance from that centre, determine 
the velocity and time. 

Let fx = the acceleration at a unit's distance from the centre 
of force ; 

x = the varying, and c the initial, distance of the body from 
that centre ; 

then XfM = the acceleration at the distance x. 

Here s=c — x: r.v = — = : (1) 

dt dt' K } 

and X/x = a = — f = ^ • (2) 

r d£ 2 d£ 2 v 7 

Multiplying (2) by —da;, we obtain 
—d(— ) = — fjixdx; 



dt \dt 

■ •••(fjM=-^ +c - 

Since « = when x = c, = /xc 2 ; 



da; 



... v = -^ = V/.(c 2 -a^); (3) 

tit 

,\ £ = /a~5cos -1 -. (4) 

c 



78 SUCCESSIVE INTEGRATION. 

(7=0, since t = when x = c. 

If in (3) and (4) we make x = 0, we have 

v = cV/x, the velocity at the centre of force; 
and t = ^7TfA~*, f 7T/x~5, etc. 

Hence the time required for the body to reach the centre of 
force is independent of its initial distance from that centre. 

Below the surface of the earth, the acceleration due to gravity 
varies as the distance from its centre. Hence, from (3) we learn 
that if a body could pass freely through the earth, it would fall 
with an increasing velocity from the surface to the centre, from 
which it would move on with a decreasing velocity, until it 
reached the surface on the opposite side. It would then return 
to its first position, and thus move to and fro. 

The acceleration due to gravity at the surface of the earth 
being g } and the radius being r, we have in this case, 

a- g • 
r 

.'.v — r V/x = V#r, the velocity at the centre ; 

a o, -i \~>' q -lA-ia (20919360 

and 2£ = 7nx 5 = tt\ - == o.l416^v sec. 

r \g \ 32.17 

= 42 min. 13.4 sec, 



which is the time that would be required for a body to fall 
through the earth. 

2. Assuming that the acceleration of a falling body above the 
surface of the earth varies inversely as the square of its distance 
from the earth's centre, find the velocity and time. 

Let x = the varying, and c the initial, distance of the body 
from the earth's centre ; 

r = the radius of the earth ; 

g = the acceleration due to gravity at its surface ; 

a = the acceleration due to gravity at the distance x. 



PROBLEMS IN MECHANICS. 



79 



Then s — c — x ; and, from the law of fall, we have 

a : g : : r 2 : x 2 ; 

_ gr 2 _ _ drx_ 
x 2 dt 2 

Multiplying (1) by dec, and integrating, we have 

v 



9 N l/l IV 



da; 
dZ 



(1) 



(2) 



If c = oo ; that is, if the body fall from an infinite distance 
to the earth, we have from (2) , when x = i\ 

v = V'2gr. 
Since g = 321 ft., and r = 3962 miles, we have 

/_64£ 3962V=6.95-h miles. 
V5280 J 

Hence, the maximum velocity with which a falling body can 
reach the earth is less than seven miles per second. 

From (2) , we have 

dx /rt _.„a< i fcx — x 2 \--_ f2gr 2 \h (cx — x 2 )? m 



^c — 2x 
2'V- 



= -('2 9 r 2 )^ 

V2^-v v^r^ 2 V 2 ^7 

( ca . _ a*) | _ £ vers- 1 — "1 + (7, 



dx; 






c 

2gS 



Since £= when x = c, C — ^c-rrl -) ; 



£ = 



2^ 



s y 



c# — ar)5 vers x p9 C7r 

y 2 c 2 



(3) 



3. Assuming that r, the radius of the earth, is 3962 miles; 
that the sun is 24,000 r distant from the earth; and that the 
moon is 60 r distant ; find the time that it would take a body to 
fall from the moon to the earth, and the velocity, at the earth's 



80 



SUCCESSIVE INTEGRATION. 



surface, of a bod}' falling from the sun. The attraction of the 
moon and sun, and the resistance of any medium, are not to be 
considered. 

4. A body falls in the air by the force of gravity ; the resis- 
tance of the air varying as the square of the velocity, determine 
the velocity on the hypothesis that the force of gravity is 
constant. 

Let fx = the resistance when the velocity is unity ; 

and t = the time of falling through the distance s. 



Then 
and 
Hence 



that is, ■ — - 



(7 \ o 
— j = the resistance of the air for any velocity ; 

g = the acceleration downward due to gravity alone. 

the actual acceleration downward ; 

ds\ 2 



fds 



dt 2 9 'V 



d(*± 

\dt 

jjl dt 



— I — - ) , or u dt 
V 



'ds\ 2 
dt) 



g_fds} 2 ' 
(x \dt 



(1) 

(2) 



Observing that the second member of (2) is of the form 
- -, and integrating, we have 



ar 



' ah 



9 h + P* 



Ag. 



l*M 



ds 
di 



(3) 



ds 



C= ; since t = when — |~= v~| = 0. 
dt L J 



From (3) , by principles of logarithms, we obtain 

e 2ty/w . 



i . ids 
a* + /x5 • — 
* ■: dt 



i ids 
q* — u* — 
U h dt 



PROBLEMS IN MECHANICS. 81 

ds fq\\e 2t ^9-l 



v. 

dt \fij e 2*vV<7 -fl 

Hence, as t increases, v rapidly approaches the constant value 

g 

P. 

5. A bod}' is projected with a velocity v into a medium which 
resists as the square of the velocity ; determine the velocity and 
distance after t seconds. 

Let fx = the resistance of the medium when the velocity 

is unity ; 

and s = the distance passed over in t seconds. 

/ds V 
Then «,[ — ] = the resistance for any velocity. 

\dtj 

rj d 2 s /ds\ 2 \dt) 

Hence — — = — a I — , or — i — L = — ads: 
dt 2 r \dt) ds r 

dt 

ds 
.-. log — = — fis + G = — ps + log v . 
dt 

ds 
C — log v , since — = v when s — 0. 

Hence — fis = log -| - log v = log f-± -*- v j ; 



ds „„ 


ds v 


i- v = e-^ s , 


or — = — ^ 


dt 


eft e^ s 



(1) 



Hence, the velocity decreases rapidly, but becomes zero only 
when s = ex. 

Integrating (1), and solving the resulting equation for s, we 
obtain 

s = -log(/x/y £ +1) • 



82 



SUCCESSIVE INTEGRATION. 



6. A body slides without friction down a given curve; re- 
quired the velocity it acquires under the influence of gravit}\ 

Let mn be the given curve, pa a 
tangent at any point p, and pa = ds ; 
then — pd = dy, and the acceleration 
caused b} T gravity at p equals g cos <f>, 
in which <f> = dpa (§ 70, Ex. 2). 

dt 2 
ds 




.'. — - — a cos ct> = 



U ds 



Fig. 23. 



-d(—)=-2gdy; 
dt \dt) y y ' 



[=v^ = -2gy + C. 



If y be the ordinate of the starting-point on the curve, v = 
when y = y , and C = 2 gry . 



• '.v=V2g(y — y). 



When ?/ = 0, v=v2gy , which is the velocity that the body 
would acquire in falling the vertical distance y (§ 70, Ex. 1). 
Hence, whatever be the curve down which, from any point p, a 
body slides without friction, it has the same velocity when it 
reaches the line ox. 



7. The base of a cycloid is horizontal, and its vertex is 
downward; find the time of descent of a heavy body from any 

point on the curve to. its 
vertex. 

Let the vertex o be the 
origin of coordinates, y 
the ordinate of the start- 
ing-point p, and s the 
length of the curve reck- 




oned from that point, 
have 



Then, from the previous problem, we 



v = V2g{y — y) = 



"111 
dt 



dt = 



ds 



V2^(y -y) 



(1) 



PROBLEMS IN MECHANICS. 83 

Since ds is positive, and dy negative, 

The equation of the cycloid referred to the axes ox and oy is 



x = r vers 1 - + V 2 ry — y 2 . 
r 

dx _ /2 r — y \g . 

■••*-(S +l ) ldir --S)V (2) 

From equations (1) and (2), we have 

<& = _ f 2r \ h d y = _ f r \ k d y . 

U / -\/2g(y -y) \9j -\/y y-f 

* = - fevers- 52 + (7. 



2/o 
Since £ = when ?/ = y ? 

?=('-Yvers- 1 2 = 7iY r 



W \ 2/o 



£ = 7r( — )", when y = 0. 



Hence, the time required to reach the lowest point o will be 

the same, from whatever point on om the body starts. Hence, 

if a pendulum swings in the arc of a cycloid, the time required 

Ir 
for one oscillation is 2 7ta J — The time of an oscillation being 

independent of the length of the arc, the cycloidal pendulum is 
isochronal. 



84 



SUCCESSIVE INTEGRATION. 



Cor. To find the time of descent along any other curve, we 
would obtain from its equation the value of c?s, substitute this 
in equation (1), and integrate between the proper limits. 

8. To find the length and equation of the Catenary. 

Let nom represent the form assumed by a chain, or perfectly 

flexible cord, of uniform section and density, when suspended 
from any two fixed points m and n ; 
then is nom a catenary. Let o, the 
lowest point, be taken as the origin. 
Let s denote the length of any arc ob ; 
then, if p be the weight of a unit of 
length of the cord or chain, the load 
suspended, or the vertical tension, at 
b is sp. Let the horizontal tension be 
ap, the weight of a units of length of 

the chain. Let bd be a tangent at b ; then, if bd represent the 

tension of the chain at b, be and ed will represent respectively 

its horizontal and its vertical tension at b. 



V 1 


> 


>± 


o 




V 


<y 




Fig 


. 25. 





TT dy ed sp s 

Hence, -?- = — = — = — 
dx be ap a 



(1) 



a[_ dxj 



•/: 



Vcfc 



dx 2 



dx 



dx 
ds 



Va 2 + 



ds 



Va 2 + s 2 
Since x = when s = 0, C= — aloga 



= o log(s + Vo 2 + s 2 ) + O. 



(2) 



Solving (2) for s, we have, for the length of the curve meas- 
ured from o, 



s = -(e a — e a ), 



(3) 



PROBLEMS IN MECHANICS. 85 

To find its equation, we have, from (1) and (3), 

-, XX 

dx 

a * -* 

Since y = when x = 0, C = — «. 

X X 

is the equation of the catenary referred to ox and oy. 

If o'o = a, and the curve be referred to the axes o'x' and 
o'y, its equation will evidently be 

X X 

a , — . --v 



CHAPTER VI. 

INDETERMINATE FORMS. 

77. When, for any particular value of its variable, a function 
assumes any one of the indeterminate forms, 

-, — , • 00, 00 - 00, 0°, 00°, 1 ±0 °, 

<*> 

the function, in the usual sense, has no value for this value of 
its variable. What we call the value of the function for this 
value of its variable is the limit which the function approaches, 
as the variable approaches this particular value as its limit. 

Often, when a function assumes an indeterminate form, its 
value may be found by algebraic methods. 



1 . Evaluate 



X s — 



In general, 

X 3 — 1 x 2 -\-x-\-l 



Examples. 
; that is, find 



limit [" or 3 — l "| 



x 2 -! 



x + l 



limit 
x=l 



ar»-r i 
j? - 1 J 



limit 
x=l 



'x 2 + x + 1" 
x + l 



2. Evaluate ( x ~ a }\ 

(x 2 — a 2 ) * 

In general, 

(x — a)* 



__ (x — a)& t 
(x 2 — a 2 )i (x — a)& (x-\- a)i (aj-f-a)i' 



(x— a)& 



. limit 
* x=a 



(x — a) 3 
(x 2 -a 2 )i_ 



limit 
x^= a 



(x—a)h 

(x + a)i 



= 0. 



EVALUATION BY DIFFERENTIATION. 87 

V2a 



3. Evaluate (a 2 -x*)k+ (g-x) 
(a-x)h+(a s -x i )h 



Ans. 



1+«V3 



Evaluation by Differentiation. 



78. To evaluate *& ,ori^i 

</>(a) <£(x) 



when it assumes the form 



and 



limit 
Ax=0 



limit 

limit 
A.r = 



~f(x + As) -f(x) 
Aa; 

>(a? + Aa;)-^(a;) 
Aa; 

7(a? + Aa-)-/(a;) - 
<H» + Aa;) -<£(») 



=/'(»), Ex. 1. p. 39. 






(i) 



Substituting a for a; in (1), and remembering that /(a) and 
<£(«) are each 0. we have 



I 
Az 



imit [" /(« + Aa;) 
* = °|_<£( a + Aa; )_ 






If /'(a) = 0, and <f>'(a) is not 0, 






= 0. 



If/'(«) is not 0. and <£'(<*) = 0, 4^7 = °°- 



K/'(a) = 0,and«£'(a) = 0, j-^ 

9 V a ) 



-' ^ '' also assumes the indeter- 



minate form - • Applying to it the preceding process of reason- 



ing, we have 



f'(a) = f"(a) 
+' (a) 4>"(a) 







If this also assumes the form - , we pass to the next deriva- 
tives, and so on, until we obtain a fraction both of whose terms 
do not equal 0. 



88 



INDETERMINATE FORMS. 



Examples. 



Evaluate 
1. 



2. 



log s i . 
aj — lji 

log x~\ _ . . logon _ x 
x — lji""0.' "a? — lj! - 1_ 

1 — cos x~\ 

~~ "S Jo" 



= 1. 



§78. 



1 — cos a; 

ar 2 



= 
o 



1 — cos af| _ sin cc" 1 
a? J ~^~_ 



_ cos a? 



3. 



a; -1 1, 
tf\-lji 

1' 



= 1 

o~2' 



a; 

e 1 — 



2. 



since 
a? — sin a? 



! 1 



6. «^£1 
a; J 

- a? — sin -1 af| 
sin 3 a; J 

79. To evaluate f W 



log 



, w7ien ££ assumes the form — 
</>(a) oo 



limit r/(g)l _ limit 
x = a \J>(x)\ x = a 



1 

l 

LA*) J 


_ limit 
x = a 


d 

dx 


1 ~] 
J>(aj)_ 


± 

dx 


-/(*)JJ 



limit f JM1 = limit f L/MT iM~] 

*=«]_*(») J *- a L[*o») ]*'/'(*) J 



(1) 



EVALUATION BY DIFFERENTIATION. 



89 



Since, when the limits of two variables are equal, the limits 
of their equimultiples are equal; we have from (1), by multi- 
4>{x) f{x) 



plying by 



f{x) *'(■)' 



limit 
x~a 



' /(%) ' 
.*(*). 



limit 
x = a 






or /W = ZM 

</><» </>'(a) 



Cor. From § 72, Cor., it follows that, if a function assumes 
the form — for a finite value of the variable, all the functions 

00 

obtained by the formula given above will also assume the form 

— Hence, to evaluate the function, it is necessarv to trans- 

oo 

form the original function, or some one of the derived functions, 

so that it will not assume the form — for this finite value of the 

00 

variable. 



Examples. 



Evaluate 



1. 



logo? ~ 




cosec x 





logo; ~ 




cosec x 





— sin 2 a;' 



00 > < log X 



X COSiC 



» cosec x 

— 2 sin x cos x~ 

cos x — x sin x 



cosec £ cotcc 



.-! = »• 



logon 

x° I 



Ans. 0. 



4. 



logo;" 



Ans. 0. 



3. 



tana? 
tan 3 x 



80. The forms 0-oo and oo — oo. Functions of x that assume 
the form • oo or oo — oo for a particular value of x, may be so 



90 



INDETERMINATE FORMS. 



transformed that they will assume the form - or — for the same 

value of x. Hence they can be evaluated by the previous 
methods. 



Evaluate 

1. 2* sin — 

2* 



Examples, 



. a 
sin — 

a 2 r 



Since 2 X sin — = 

2* 2~ x 



and 



= a cos — 



= a : 



2. (1 — a?)tan — 

V ; 2 



Since (1 — x) tan— == 

cot 



2 X sin — = a. 

2* 

_l 00 



and 



l-x ~ 

, irX 

cot 

2 



it oirX 

- cosec z — 

1 2 2 J] 



(1 — x) tan 



ttX 



= 2 c 
i t 



3. [sec x — tan x] ^ . 
Since sec x — tan x = 



sin x 1 — sin x 



cosx cos a? COS a 



, 1 — sin x ~\ _ cos# 

cos x J i„. sin a;^ 

[_log# logajjj 



= 0; .*. [sec x — tan a;]| ir = 0. 



Arts. 

2 



-1 






EVALUATION BY DIFFERENTIATION. 



91 



81. The Forms 0°, a>°, and l ±x . When, for any value of x, a 

function of x assumes one of the forms, 0°, oo°, or l ±x , its loga- 
rithm assumes the form, ±0«o>, and the function is evaluated 
by evaluating its logarithm for this particular value of x. 



1. Evaluate af] . 
Since log of = x log x 



Examples. 

x 



\ogx 



and 









2 


X 


-(logxf 


2 log x 


X 


1 


1 


1 


1 


log X 


o *£ 


o *£ _ 


o «*r 



2x 



__ 



.\logaf] = 0, or af] = 1. 

2. a^^o. 

los" X 
Since loga sin:r = sin a 102^ = — - — , 

cosec x 



and 



logo; T 

cosec x_\ 



— cosec x cot x 

— 2 sin a; cos x~ 



— sur a; 

£COS# 



= 0; 



cos a? — x sin cc 
.♦. loga sin *] = 0, or a sin *] = 1. 

3. sina sin:,: ] . Ans. 1. 



= 0: 



4. sino; tan - r ]^. ^l?is. 1. 



82. Compound Indeterminate Forms. When the given func- 
tion can be resolved into factors, some or each of which assumes 
the indeterminate form, each factor may be evaluated separately. 

Thus, if the function be ^-^ 



? — l) tan 2 aT 1 



we have 



92 



INDETERMINATE EOHMS. 



since 



(e»- 


1) tan 2 aT 


/'tana 


x :i 


~\ X 


tan af 

X 


1 a e ~ l 
= 1, and 

o X 



= 1 



= 1, 



Examples. 



tan a; — x 



x — sin a? 



= 2. 



2. 



afl 1 
Jo 6' 



4. 



5. 



x — sin 
x 3 

e x — e~ x ~ |_ 
log(l + as)J 

e x — e _x — 2 a; 



(e*-l) ; 



Jo 3 



f]. 



1 — sin x + co s 
sin a; + cos x — lji^ 

1.1, as-e*J 



GO. 



G. 



a: 2 + 2 cos a? — 2" 



as 3 — 3a;-}-2 
a; 4 — 6x 2 +8x — 3 



12 



8. 



<x sinj •' — a 
log sin as 



= a log a. 



9. xi- 



10. (cos-mas)* L=l. 



12. [as tanas — |-7r sec as] i^ = —1. 



13. 



cos as — cos at 

o-x 2 d £>-a 2 d 



e u u sm a i 



2a 



<-> 



= e fl 



83-. Evaluation of Derivatives of Implicit Functions. If an 

equation containing as and y is solved for ?/, y is an explicit func- 
tion of as ; if it is not so solved, y is an implicit function of x. 



EXAMPLES. 



93 



When y is an implicit function of a?, its derivative, though 
containing both x and y, is a function of x. Hence, when the 
derivative assumes an indeterminate form for particular values 
of x and y, it can be evaluated by the previous methods. 



Examples. 
1. Find the slope of a 2 y 2 — a 2 x 2 — x 4 = at (0, 0). 

TT dv 2a 2 x+4:X 3 , n 

Here — = = -, when x = y = 0. 

dx 2a 2 y U 



TT dy~\ 2a 2 x-±-4:X s 

Hence — = 

dx^j 0| 



2a 2 2/ Jo,o 



2a 2 +12x 2 ~ 



2<x 



>dy 
dx 



1 

o.o dx 



dx) 



= 1, or 



dec 



= ±1. 



2. Find the slope o£y s =ax 2 — x i at (0,0), 



Here -^ 






Sy 2 

2 a — 6 x 



0,0 



62/ 



0,0 



2a — 6a; 




6»4 

da; 


? 

0,0 


2a 

o~ " °°' 


dy~ 
or -^ 
dx 



= ± 00, 



3. Find the slope of ar 3 — 3 aoy + y s = at (0, 0) . 

Ans. -^ =0 and oo. 



dx 



4. Find the slope of x 4 - a 2 xy -f b 2 y 2 = at (0, 0) . 



Ans. & 
dx 



= and — -• 



5. Find the slope of (y 2 + a^) 2 — 6 aa;y 2 — 2 aar 3 -f- a 2 ^ = at 



(0,0) and (a, 0) 



dx 



J 0,0 



± 00 



dy 
dx 



= ± 



CHAPTER VII. 
DEVELOPMENT OF FUNCTIONS IN SERIES. 

84. A Series is a succession of terms following one another 
according to some determinate law. The sum of a finite series 
is the sum of all its terms. An infinite series is one the number 
of whose terms is unlimited. 

If the sum of the first n terms of an infinite series approaches 
a definite limit as n increases indefinitely, the series is Con- 
vergent ; if not, it is Divergent. 

The limit of the sum of the first n terms of an infinite con- 
vergent series, as n increases, is called the Sum of the series. 
An infinite divergent series has no definite sum. 

85. To Develop a function is to find a series, the sum of 
which shall be equal to the function. Hence the development 
of a function is either & finite or an infinite convergent series. 

For example, by division, we obtain 

1 ~ xn = i + a; + a 2 + a 3 H \-x n ~\ 

1 —x 

\ x n 

This finite series is evidently the development of - — — for 
any value of x. 

Again, by division, we obtain 



1 + x -f x 2 + x 3 H h af- 1 + 



1 — X 1 —X 

x n 1 

Hence is the difference between and the sum of 

1 —x 1 —x 

the first n terms of the series. For x < -f 1 and >— 1, this 
difference evidently =0 as n increases, and the series is the 
development of the function. But for x> + l or < — 1, this 
difference increases numerically as n increases, and the series is 
not the development of the function. Thus, the series equals 
the function for x = J, but not for x = + 2 or — 2. 



taylor's formula. 95 

86. Taylor's Formula is a formula for developing a function 
of the sum of two variables in a series of terms arranged accord- 
ing to the ascending powers of one of the variables, with coeffi- 
cients that are functions of the other. 

A general symbol for any function of the sum of x and y is 
f(x + 2/), of which (x + y) m i \og(x + y), a x+y , sin(x + y), etc., 
are particular forms. 

87. To produce Taylor's Formula. 

We are to find the values of A, B, C, etc., when 

/( x -h y ) = A + By + Cy 2 + Dy 5 + W + • • • , 

in which A, _B, (7, i), etc., are functions of x, but independent 
of 2/, the series being finite or convergent. 

Let x' be am T value of a?, and A', B\ C", etc., the correspond- 
ing values of ^4, -B, C, etc. ; then we have 

f(x'+y) = A'+B'y + C'y 2 + V'y B + E'y i +-~; (1) 
... f( x '+ y ) = B '+ 2 Cy + 3D'y 2 + 4£y + ■-, (2) 

/'"(a'+y) = 2 • 3D'+ 2 • 3 • ±Wy + •••, (4) 

/-(^+2/) = 2.3.4^'-f---.,etc. (5) 

These equations, being true for all values of y, are true when 
y = (§ 6) ; hence we have 

/(*') = ^'» f{x') = B\ f"(x') = 2C, 

f'"(x') = \3D', f"(x') = \±E', etc. 

Solving these equations for A', B', C", etc., we have 

^'=A«0, B'=f(x'), 0'=£%fK 

D'^r^l, E'-^M, etc. 



96 DEVELOPMENT OF FUNCTIONS IN SERIES. 

Substituting these values in (1), we obtain 

f(x<+y) =f{x<) +f'(x')y +/»(«') g +f"(x') £- 

+/"(*') j| + " (6) 

Since the coefficients in (6) are equal to f(x), f'(x), /"(#), 
etc., for x = x', and since x' is any value of sb, we have, in 
general, 

f(x + y) =f{x) +f'(x)y +f"(x) £ +/'"(*) £ 

+/"(*) j£_+- < A > 

This development of f(x-\-y) was first published in 1715 by 
Dr. Brook Taylor, from whom it is named. 

88. When x'= 0, equation (6) of § 87 becomes 

Ay) =/(o) +/'(0)2/ +/"(«) =£ +/'"(0) =£ 

+/"(0-)j| +•■• (7) 

in which /(0), /'(0), etc., are the values of f(y) and its suc- 
cessive derivatives when y = 0. 

Letting a? represent the variable in (7), we obtain 
f(x)=f(0)+f'(0)x+f"(0)^ + f"'(0)^. 

+ /"(0)g+... (B) 

Equation (B) , called Maclaurin's Formula, is a formula for 
developing a function of a single variable in a series of terms 
arranged according to the ascending powers of that variable, 
with constant coefficients. 



THE BINOMIAL THEOREM. 97 

The completion* of Taylor's and Maclaurin's formulas will be 
deferred until we have applied them to the development of a 
few functions. 

89. To develop (x -f-y) 111 , or to deduce the Binomial Theorem. 
Here f(x + y) = (x + y) m ; ,:f(x) = x'% 

f(x) = mx m -\ f"(x) = m{m — l)x m ~\ 

f'"(x) = m(m - 1) (m - 2)x m ~\ etc. 

Substituting these values in Taylor's formula, we have 

(x + y) m = x m + mx m ~ x y + m ( m ~ 1 ) a*-*y 

. m(m — 1) (m — 2) m _ 3 3 . 
H — ^ -^ J -x m y + ... 

90. To develop log & (x -f y) . 

Here f(x + y) = log, (x + y) ; .-. /(a?) = log a x, 

/'(*) = ?, f"(x) = -% f"(x) = *£, etc. 
a; ar x i 

Substituting these values in Taylor's formula, we have 

log. (a? + y) = log a « + w/2 - X + X - JL + ••• 

\x 2ar 3ar 4 ar 

which is the logarithmic series. 

Cor. If 05= 1, and m = l, we have 

log(l + 2/)=^-£ + ^-^+..., 

which is the Naperian logarithmic series. 

* The series obtained by applying Taylor's or Maclaurin's formula, as 
given above, to any given function may or may not be the development of 
that function. Their complete forms, however, enable us to determine what 
functions can be developed by them- 



98 



DEVELOPMENT OF FUNCTIONS IN SERIES. 



91. To develop a x+y . 

Here f(x) — a*, f'(%) = a x log a, etc. ; 



.-. a^ y = ar 



V 



2?r 



f 



1 + log a ^ + (log a) 2 ^- + (log a) 8 %- . 



92. To develop (a + x) m fo/ Maclaurin's Formula. 
Here /(#) = («+^) m ; •'• /(0) = « m . 

/'(a;) = m(a+a;) m - 1 ; /. /'(0) = ma^ 1 . 

/"(a?) = m(m-l) (a+a;) m - 2 ; ."•'. /"(0) = m(m-l)a w - 2 . 
etc. etc. 



Substituting these values in Maclaurin's formula, we have 
(a + x) 



m . m _i . m(m— 1) ™_ 2 9 , 



93. To develop sin x. 
Here f(x) = sin x ; 
f'(x) = cos a;; 
f"(x)= — sin a?; 
f'"(x) = — cos a?; 
/ IV (a?) = sin a;; 
f v (x) = cos a?; 
etc. 



11 

.'. /(0) = 0. 
/. /'(0) = 1. 

: v/"(o) = a. 
.•./'"■(.o) = -.i. 

.-./~(0) = 0. 

■•■ / y (0) = i. 

etc. 



Substituting these values in the formula, we have 

/yv> /y*> ,yti /^,J 

sin x = a? — • h , h ; • • • 

II li 11 Li 

94. To develop cosx. 

1w ry& /y»6 /yo 

[2_ |4_ [6_ [8_ 

This result could be obtained by differentiating the value of 
sin a; found in § 93. 



THE LOGARITHMIC SERIES. 99 

95. To develop a x . 
a^l+loga^4-(loga) 2 ^+(loga) 3 ^ + ... 5 

which is the exponential series. 

Cor. 1. If a = e, the Naperian base, we have 

Cor. 2. Putting a?= 1, we have 

e=l+1+ i + ii + i + i + - 

Hence e =2.718281 + . 

96. To develop log,{\ + x) . 

\og a (l + x) = m(x-^ + y-f +f--} (!) 

which is the common logarithmic series. 
If m = 1 , we have 

log(l + x) = x - — + — - — + — , (1') 

oV; 23 45 W 

which is the Naperian logarithmic series. 

In § 104, this development is proved to hold only for values 
of x between —1 and -j-1 ; hence, in this form, it is useless for 
the computation of Naperian logarithms of numbers greater 
than 2. We therefore proceed to adapt the Naperian logarith- 
mic series to the computation of logarithms. 

Substituting — x for x in (1'), we have 

/y»* /yi*> /y** /y»0 

log(l - x) = - x - — - — - — - — (2) 

&v } 2 3 4 5 V 

Subtracting (2) from (1'), we have 

l0g(l + ^)-l0g(l-^) = 2^ + ^ + y+y4---)-(3) 



100 DEVELOPMENT OE FUNCTIONS IN SERIES. 

I I _i_ x z 4-1 

Let x — ; then = - ; and, for any positive 

u Z -\~ I I — X Z 

value of z, x< 1. 

Hence log(l + a?) — log(l — a?) = log(2 + 1) — log z. 
Substituting these values in (3), we have 
log (z + 1) -logz 

2z+l + 3(2z+l) 3+ 5(2z+l) 5+ '"/ 

Equation (4) is true for any positive value of z ; and, since 
the series converges rapidly, log (z + 1) can be readily computed 
when \ogz is known. 

Putting z = 1 in (4) , we obtain 

log2 = 2^+A, + ^ + 



^3 3-3 3 5-3 5 7-3 7 
Summing six terms of this series, we find 

log 2 = 0.693147 + . 
Putting 2 = 2 in (4) , we have 

108^^2 + 2(1 + ^ + ^ + ^ + ...) 

= 1.098612 + . 
log 4 = 2 log 2 = 1 .386294+ . 

Putting 2 = 4, we obtain 

lo g5 = log4+2(i + 3 -^ + 5 -L. 5 + 7 -^+...) 

= 1.6094379 + . 
log 10 = log 5 + log 2 = 2.302585 + . 

In this way we can compute the Naperian logarithms of all 
numbers. 

Cor. 1. Letting m and m' be the moduli of two sj'stems of 
logarithms whose bases are a and a', from (1) we evidently 
ha ve log a (l + s) = m T (5) 

log a -(l + a) m' 

in which # lies between —1 and +1. 



THE LOGARITHMIC SERIES. 101 

To prove that the principle in (5) is true for all numbers, let 
log a (1 + x) = tt, and log a - ( 1 + x) = w ; 

w 

then a M = (l'+ x) = a' w , or a' = a w ; (G) 

and log.(l + s) = JL (7 ) 

log a .(l + a?) lu 

Again, y being any number, let 

\og a y = 2, and log,,-// = w ; 

z 

then a? = y = a' v , or a' = a*; (8) 

and kfeJUL (9 ) 

From (6) and (8), 

a w = «% or — =— (10) 

to 1' 

From (7), (9), (10), and (5), we have 

log a y = log a (1+x) = m ^ 

log a .?/ log a .(l + a;) m' 

Hence, f/^e logarithms of the same number in different systems 
are proportional to the moduli of those systems. 

Cor. 2. If, in (11) of Cor. 1, we let a'= e, we have 
m'= 1, 
and \og a y = mlogy. (12) 

Hence, the logarithm of a number in any system is equal to 
the Naperian logarithm of the same number multiplied by the 
modulus of that system. 

Cor. 3. If, in (12) of Cor. 2, y= a, we have 

1 

m = 

log a 

Hence, the modulus of any system of logarithms is the recipro- 
cal of the Naperian logarithm of its base. 



102 DEVELOPMENT OF FUNCTIONS IN SERIES. 

In the Common system, a = 10 ; 

hence m — = =.434294 + . 

log 10 2.302585 T 



97. Taylor's formula evidently fails to develop f(x -f- y) for 
x = a, if /(a) or /"(a) is infinite, while n remains finite; while 
Maclaurin's fails to develop f(x) for any value of x, if /(0) or 
f n (0) is infinite for n finite. 

For example, by Taylor's formula, we have 

* 

V y) K ■ 2{x-b)h 8(x-b)% K ) 

Whence =6, (1) becomes 

•y/y = go — oo -f- • • • 
Hence the formula fails to develop (x — b + y)* for x = b. 

By Maclaurin's formula, we have 

log£=— oo + oo — oo -f- .-• 

Hence Maclaurin's formula fails to develop log x for any value 
of ct*. 

98. To complete Taylor's and Maclaurin's formulas so that 
they shall enable us to determine what functions can be devel- 
oped by them, we need the following lemma : 

Lemma. If f (x) is continuous betiveen x = a and x = b, and 
iff (a) = f (b) = 0, then f'(x), if continuous, must equal zero for 
some value of x between a and b. 

For, if f(x) is continuous, and f(a) =/(&) = ; then, as x 
changes from a to 5, f(x) must first increase, and then decrease ; 
or first decrease, and then increase; hence f'(x) must change 
from + to — . or from — to +, and therefore, if continuous, 
pass through 0. 



taylor's and maclattrin's formulas. 103 

99. Completion of Taylor's and Maclaurin's Formulas. To be 

the development of f(x-\-y), the series in Taylor's formula must 
be finite or infinite and convergent (§ 85). 

Let P^~ be the difference between f(x + y) and the sum of 

\n 

the first n terms of the series ; then we have 

/(* + y) =A») +/'(*) \ +/"(f) if +/"'(<») jf 

+ ... + /«-i (a; )jrL + .pj£. ( i) 

I ?i — 1 \n 

We proceed to find the value of P. 

Letting y = X — x in (1), and transposing, we have 
f(X) -f(x) -f{x) *=* -f'(x) {X ~ XY 

-p ( X ~^ n = 0. (2) 

[£. 

Let i^(z) represent the function of 2 obtained by substituting 
2 for x in the first member of (2) ; then 

F{z) =/(X) -/(*) -/'(*) *=* -/"(*) i~^ 

_..._ r - I( ,)(^z£lr!_p(^I'. (3 ) 
|n — 1 \n 

Substituting X for z in (3) , we obtain F{X) = 0. 

From (2) we see that the right-hand member of (3) is for 
z = x ; hence, by substituting cc for z in (3) , we obtain F(x) =0. 

Differentiating (3) to obtain F'(z), we find that the terms of 
the second member destroy each other in pairs, with the excep- 
tion of the last two, and obtain 



104 DEVELOPMENT OF FUNCTIONS IN SERIES. 

F'(z) = - ( X ~ Z " }n ~ 1 f n (z) + ( X - Z ) n ~ 1 p. 

I 71 - 1 \n-l 

Whence, P=f n (z) when F'(z)=0. Since F(z) = when 
z = X, and also when z = x, F'(z) = for some value of z 
between X and x (§ 98) . Now, by giving to some value be- 
tween and +1, any value between x and X can evidently be 
represented by x -f 0(X — x) . 

Hence, P=f n [x 4- 0(X- a;)] =f*(x + By). 
Substituting this value of P in (1), we have 

\n — 1 [w_ 

which is one complete form of Taylor's formula. 

Cor. Letting x = 0, and putting a? for ?/, we have 

/(*) =/(<>) +/'(<>)* +/"(0) £ +/"'(0) j£ + ... 

+/"- 1 (0) r ^l+/»(^)^, 

I ?i — 1 \n 

which is one complete form of Maclaurin's formula. 

100. If we had let P x y be the difference between /( x -\-y) and 
the sum of the first n terms of the series in Taylor's formula, 
we should have found 

\n — 1 
Hence a second complete form of Taylor's formula is 

A* + v) =/(*) +/» f +/"(«) jf + - 

n — 1 \n — I 



Taylor's and maclaurin's formulas. 105 

Cok. 1. The corresponding complete form of Maclaurin's 
formula is evidently 

j\x) =/(0) +f'{0)x +/"(0)^ + ...+/-i(0) J£L 

|J_ [ n-1 

I 71 — 1 

All that we know of in any of these formulas is that its 
value lies between and +1. 

Cor. 2. If, on applying to a given function any one of these 
completed formulas, the last term becomes 0, or approaches 
as its limit, as n increases, the formula evidently develops this 
function ; if not, the formula fails. 

101. Since f- = 2- . -£ ; and since - is very small when y 

\n_ n I n — 1 n 

is finite, and n is very large ; each value of — is a very small 

llL 
part of its preceding value. 

y n 
Hence , — = 0, ivhen y is finite and n increases indefinitely. 

Cor. 7/*f n (x) does not become infinite with n, Taylor's and 
Maclaurin' s formulas give the true development of f(x-fy) and 
f(x) respectively. 

102. To prove that Maclaurin's formula develops a x . 
Here f n (x) = (log a) n a x ; . \ f n (0 x) = (log a) n a 9x , 

and rW*)-= {xl ° gaY a. 9 '. 

\n \n 

Since a Qx is finite, and ^ — -s — L = as n increases indefi- 

\n 
nitely (§ 101) ; L ~ 



(g lo £ a ) 



a 0x = O 

n 



as n increases, and the formula develops the function (§ 100. 
Cor. 2) . 



106 DEVELOPMENT OF FUNCTIONS IN SERIES. 

103. To prove that Maclaurin's formula develops sin± and 
cosx. 

The nth derivative of each of these functions is finite, how- 
ever great n may be ; hence Maclaurin's formula develops both 
of them (§ 101, Cor.). 

104. To determine for what values of x Maclaurin's formula 
develops log(l + x.). ' 

The formula gives (§ 96), 

x 2 . a? x* , , (-i)— "af- 1 

. -J _ . -+_ ... -X- -^ £ . 



l g(l + aO=7-^-+lT--T-+--- + 



12 3 4 n-1 

n 

The ratio of the nth term to the term before it evidently 
approaches — x as n increases. Hence, if x is numerically 
greater than 1, the series is divergent, and cannot be the devel- 
opment of log(l + x). We need, therefore, to examine the 
value of the last term of the formula only for values of x be- 
tween — 1 and +1. 

(_l)-iu_l 

.\f"($x)— = (- 1 )"" 1 



\n n \l + 9 x J 

( — \\n-l / x \n 

For values of x between and -h 1 ? - — and f ) 

n \l + 0xj 

each approaches as n increases ; hence the formula develops 

log (1 -f- x) for these values of x. 

When x lies between and — 1 , let x 1 represent its absolute 
value ; then X = — x 1 , and log(l -+- x) = log(l — a^) . 

Using the second form of the formula, we have, numerically, 

ri6Xl) (i-tf)-V = a-ey-w 

\n — l {l — 6x x ) n 

-Ox^- 1 



1-OxJ l-$x l 



taylor's and maclattrix's formulas. 107 



For values of x 1 between and -f-1, — — — is finite, and 

l — 0x 1 

zl Li approaches as n increases. The formula there- 

\l-0xj 

fore develops log (l-\-x) when x lies between and — 1. 

Hence, Maelaurin's formula develops log(l + x) for values of 

x between — 1 and -f-1. 

105. To determine for what values of x Maelaurin's formula 

develops (l-f-x) m . 
The formula gives 

a i \m -1 r i W (ill 1) a . 
-f- x) m = 1 -f- mx -\ ^— L xr -j 

, m(m — 1) ••• (m — ?? + 2) 

H - ^ :L -¥-x n ~ 1 . 

n —1 

x n 
If m is a positive whole number, f n (0x) — =0 when ?i = m + l ; 

[n_ 

hence, in this case, Maelaurin's formula develops (l-{-x) m in a 

finite series of m-\-l terms. 

If m is negative or fractional, the series is infinite. The ratio 

of the (n + l)th term to the nth is — ~ 1 a\ which approaches 

n 

— x as n increases. The series therefore is divergent, and can- 
not equal (1 + ®) m when x is numerically greater than 1. Hence 
we need examine the value of the last term of the formula only 
for values of x between + 1 and — 1 . 

Here r(x) = M»i-i)-(m.- n + !)(! +x)" 

V ' (1+3!)" 

[n 
= rm(m-l). v (m^ W +l) aft l/_j_Y 

L \n ]\l+dxJ K } 

When x lies between and 1, (1 -f 0x) m is finite, 
and | ) = 0, as n increases indefinitely. 



108 DEVELOPMENT OF FUNCTIONS IN SERIES. 

An increase of 1 in the number of terms multiplies the factor 

in brackets by x, which approaches — x as n increases. 

n + 1 
Hence the last term of the formula approaches as n increases ; 

and the formula develops (l + x) m for values of x between 

and +1. 



Using the second form of the formula, we have 

-oy- 

[71-1 



f n (ex) ^ 1 ~^ nlxn 



±0xY 



= m(m-l)...( TO -Hl) f 1/ l^ y(l- 
\n-l ]\l + 6x) 1 

( 1 -4- $x} m 

For values of x between and — 1 , ^— — '— is finite ; 

1-0 

1 n \n 

approaches as n increases ; and an increase of 1 in 



l + 0x t 

the number of terms multiplies the factor in brackets by x, 

n 

which approaches — x as n increases. Hence Maclaurin's for- 
mula develops (l + x) m for all values of x between —1 and +1. 



106. The Binomial Theorem. Since (a + x) m = a m (l -f - J , 

(x\ m 
14- -) can be developed by Maclaurin's formula, when x 
aj 

is numerically less than a (§ 105) ; therefore, in this case, 

(a + x) m = a m -f ma m - x x + m (^ -1) a m ~ 2 x 2 + ••• (1) 

For like reason, when x is numerically greater than a, 

{x + a) m = x m + mx^a + m <>-l) SB m-2 a * + ... (2) 

11. 

Hence one, and only one, form of the development of (a + x) m 
holds for any set of values of a and x. 



THE COMPUTATION OF 77. 109 

107. To develop fcm -1 x, and find the value of V. 

When # lies between — 1 and + 1 •> 

— —, ■ = (l+^)- 1 = l-aP+x'-^+^ ; §105. 

1 -f-ar 

.'. | — — 2 = I dx— I x 2 dx + j x' 4 <ix — I # 6 cfcc 
-f- I x?dx — • • • 
Hence, if a; is numerically less than 1, 

tan" 1 ^ = a; - — + — - — + — (1) 

3 o / 9 

(7=0, since tan _1 cc = when x = 0. 

If we put <B=-ij-, equation (1) becomes 

tan-Ji = -=JYl-i + ^-^ + -..V 
\3 6 \3V 9 45 189 / ' 

.•. 7r = 2V3( / 'l-i + ^ — + .••] = 3.141592 + . 

^ 9 45 189 J 

108. To develop sin^x, and find the value of V. 

. -, . a; 3 , 1-3^ . 1.3.5a? 7 , 

sin ± a! = aH , 

2-3 2-4-5 2.4.6.7 

when x lies between — 1 and + 1 . 

,.- = Yi+-+— +— +■ 

6 2 V 24 640 7168 
or, 7T = 3.141592 + . 

It was by means of this series that Sir Isaac Newton com- 
puted the value of 71-, 



110 



DEVELOPMENT OF FUNCTIONS IN SERIES. 



109. To prove geometrically that f (a + h) = f (a) + h f ' (a-f h) , 
f (x) and f (x) being continuous and finite be- 
tween x =a and x = a -f- h. 

Let pp'p" be the locus of y =f(x) , a = oa, 
and h = an ; then / (a) = ap = nm, and 
f(a + h) = np". The curve must be parallel 
to the chord pp" at some point p' between p 
and p". Now, ob = oa -j- ab = a -+- Oh, hav- 
ing some value between and 1 ; 

hence f'(a -\-0h) = tanBDp' = tanMPp". 

.'. mp"= pm tanMPp"= hf'(a + 6h) ; 

,\f(a + h) = nm + Mp"=/(a) + hf(a + 0h). 




Examples. 



1. Develop (a 2 + 5^)5. 



A f 2 , ,,, . bx 2 b 2 x 4 . b s x 6 

Ans. (a- -\-bx-)* = a-\ -\ 

v y 2a 8a 3 16a 5 



2. Prove that tan x = x- 



af 2X 5 . 
3 15 



3. Prove that sec x= l-\ 1 -\ 

2 24 720 



4. Given f(x) = 2 a; 3 — 3 a? 2 -f- 4 a: — 3, to find the value of 
f(x + 7i) , h being a variable increment of x. 

Here /(a;) of Taylor's formula is 

2x? — 3ar4-4x — 3, and y = h; 

.-. /'(») = 6a; 2 -6a; + 4, /"(a>)= 12oj — 6, 

/"'(a;) = 12, and/ IV (a;) = 0. 

.-. f(x + 7i) = 2 a; 3 - 3 a; 2 + 4 a; - 3 + (6 a; 2 - 6 x -f- 4) ft 
+ (6a;-3)7i 2 + 2/i 3 . 

5 . Given /(aj) = 2 a; 5 - 3 a;, to find f(x + h) . 



EXAMPLES. Ill 

6. Develop sin (x -f- y) . 



sin (x -\-y) = sin x ( 1 — 



li H li / 



\ 11 11 \L 

= sin x cos 2/ + cos x sin y. §§93,94 

7. Prove that cos (x -f- y) = cos a; cos y — s'mx sin ?/. 
i ar + 4a^_31a^ + 



11 |i 

/y»^ /y»o /y»4: 

9 . Prove that log ( 1 -f- sin x) = x 1 [-••• 

&V } 2 6 12 

10. Developed*. e' inx = 1 + a? + ,— - ~ + — -— ' + 

11 II 11 11 

11. Develop 



V& 2 - cV 



12. Develop arV. #V = a; 2 + x 3 4- — 4- — + ••• 

[2_ [I 

13. Develop log(H-e-). log 2 + | + |._Jj^ + ... 

14. Prove that, if /(a;) =/( — x) , the development of f(x) 
contains only even powers of x ; while, if f(x) = — /(— x), the 
development of f(x) involves only odd powers of x. 

15. What powers of x appear in the development of since, 
and why? cos a;? tana;? sec a;? sin -1 a;? tan ^x? 



CHAPTER VIII. 
MAXIMA AND MINIMA. 

110. A maximum value of a function of a single variable is 
a value that is greater than its immediately preceding and suc- 
ceeding values. A minimum value is one that is less than its 
immediately preceding and succeeding values. 

Therefore, if we conceive x as always increasing, f(x) must 
be an increasing function immediately before, and a decreasing 
function immediately after, a maximum ; also, immediately be- 
fore a minimum, /(as)- is a decreasing, and immediately after 
an increasing function. 

Hence, f (x) is positive before and negative after a maximum 
o/f(x), and negative before and positive after a minimum. 

111. From § 110, f'(x) must change its sign as f(x) passes 
through either a maximum or a minimum. But, to change its 
sign, /'(as) must pass through or go. 

Hence, any value of x that renders f (x) a maximum or a 
minimum is a root of f (x) = or f (x) = oo. 

The converse of this theorem is not true ; that is, any root 
of /'(as) = or oo does not necessarily render /(as) a maximum 
or a minimum. These roots are simply the critical values of x, 
for each of which the function is to be examined. 

To illustrate geometrically the preceding definitions and 
Y principles, suppose a'h' to be the locus 

c of y = f(x) . Then, by definition, aa', 

VL/Tn . cc', and ee' are maxima, and bb' and dd' 

; ', \ € 

\&AW are minima of f(x) . In passing along 

I J I i H x the curve from left to right, the slope 
oa b c d e h Q f ^ e curv e f{x) is positive before, 

and negative after, a maximum ordi- 
nate ; and negative before, and positive after, a minimum 
ordinate. 



METHODS OF FINDING MAXIMA AND MINIMA. 113 

Moreover, at a point whose ordinate f(x) is a maximum or 
a minimum, the curve is either parallel or perpendicular to the 
axis of 05, and therefore f'(x) = or oo. 

At x = oh, f'(x) = Q, but hh' is neither a maximum nor a 
minimum of f(x) . 

112. Whether any one of the critical values of a; renders /(a?) 
a maximum or a minimum can be determined by one of the 
following methods : 

First Method. In this method we determine directly whether 
f'(x) changes from positive to negative, or from negative to 
positive, as x passes through a critical value. 

Let a be a critical value, and h a very small quantity. In 
/' (x) substitute a — It and a + h for x. 

If f'(a-h) is positive, and f'(a + h) negative, 

f(a) is a maximum. § 110 

If f'(a-h) is negative, and f'(a-\-h) positive, 
f(a) is a minimum. 

If f'(a — h) and f (a + h) have the same sign, 
/(«) is neither a maximum nor a minimum. 

Second Method. This method applies only to the roots of 
f'(x) = 0. Let a be a critical value of x. Developing f(x—7i) 
and f(x + h) by Taylor's formula, substituting a for x, trans- 
posing f(a) , and remembering that f'(a) = 0, we have 

/(a-70-/(a)=/^a)|-/'>)^V/-(a)^-..., (1) 

\± L£ 11 

and /(a + ZO-ZC^^/'Ca)^/''^^^/^^^--- (2) 

If h be taken very small, the sign of the second member of 
either (1) or (2) will be the same as the sign of its first term. 
Hence, if /"(«) is negative, f(a) is greater than both /(a — h) 
and /(a + 7i), and therefore a maximum; while, if /"(«) is 
positive, /(a) is less than both f(a — h) and f(a + h) , and 



114 MAXIMA AND MINIMA. 

therefore a minimum. If f"(a) is 0, and f'"(a) is not 0, 
/(a) is neither greater than both f{a — 7i) and /(a + /*) , nor 
less than both, and is therefore neither a maximum nor a mini- 
mum. If /'"(a), as well as /"(a), is 0, and/ lv (a) is negative, 
/(a) is greater than both f(a — li) and f(a + 7i) , and therefore 
a maximum; while, if / IV (a) is positive, J\a) is a minimum, 
and so on. * 

Hence, if sl is a critical value obtained from f'(x) = 0, substi- 
tute a for x in the successive derivatives of f(x). If the first 
derivative that does not reduce to is of an odd order, f (a) is 
neither a maximum nor a minimum ; but, if the first derivative 
that does not reduce to is of an even order, f (a) is a maximum 
or a minimum, according as this derivative is negative or positive. 

113. Maxima and minima occur alternately. 

Suppose that a < 6, and that f(a) and f(b) are maxima 
of f(x). When x = a + h, f(x) is decreasing; and, when 
x = b — h, f(x) is increasing, h being very small. But, in 
passing from a decreasing to an increasing state, f(x) must pass 
through a minimum. Hence, between two maxima, there must 
be at least one minimum. 

In like manner, it can be proved that between two minima 
there must be at least one maximum. 

114. The solution of problems in maxima and minima is 
sometimes facilitated by the following considerations : 

(a) Any value of x that renders c -f(x~) a maximum or a minimum, c 
being positive, renders f(x) a maximum or a minimum. 

(6) Any value of x that renders log a f(x) a maximum or a minimum, 
renders f{x) a maximum or a minimum, a being greater than unity. 

(c) Any value of x that renders f(x) a maximum or a minimum, renders 
a minimum or a maximum. 

(d) Any value of x that renders c +f(x) a maximum or a minimum, 
renders f(x) a maximum or a minimum. 

(e) Any value of x that renders f(x) positive, and a maximum or a 
minimum, renders [/(^)] n a maximum or a minimum, n being a positive 
whole number. 






EXAMPLES. 115 



Examples. 



1. Find what values of * render 4*' — 15 xr + 12* — 1 a 
maximum or a minimum. 

Here /(*) = 4* 3 - 15* 2 + 12*- 1 ; 

.\f'(x) =\2xr- 30* + 12, and /"(*) = 24a - 30 ; 

hence the roots of f'(x) = 12* 2 — 30 x + 12 = 0, which are £ and 
2, are the critical values of x (§ 111). 

But /"(i) = [24* -30], = -18, 

and f"(2) = [24* - 30] 2 = + 18 ; 

hence when * = J, the function is a maximum (§ 112), and 
when * = 2, it is a minimum. 

2. Find the maxima and minima of X s — 9 x 2 + 15 * — 3. 

Here /'(*) = 3* 2 - 18* + 15, and /"(*) = 6*- 18 ; 

therefore 5 and 1, the roots of /'(*) = 3* 2 — 18* + 15 = 0, 
are the critical values (§ 111). 

/"(5) = [6*-18] 5 = + 12, and /"(l) = -12; 

-\/(5), or [* 3 -9* 2 +15*-3] 5 , [=-28] isamin., 

and /(l)? or [jx? — 9ar +15* — 3]^ [= 4] is a max. 

Let the student construct the locus of y = * 3 — 9* 2 + 15* — 3, 
and thus exhibit these results geometrically. 

3. Examine * 3 — 3* 2 +3*+7 for maxima and minima. 
Here /'(*) = 3* 2 -6*+3, /"(*) = 6*-6, and /'"(*) =6 ; 

therefore 1 is the critical value. 

But /"(1) = [6*-6] 1 =0, and /'"(1) = 6; 

hence the function has neither a maximum nor a minimum (§112). 

4. Examine x 5 — 5 * 4 + 5 * 3 — 1 for maxima and minima. 
Here /(*) = x 5 — 5 * 4 + 5 se 3 — 1 ; 



116 MAXIMA AND MINIMA. 

.-./'(a?) = 5a? 4 -20ar 3 + 15a? 2 , /"(a?) = 20a? 3 -60a? 2 +30a?, 
and f'"(x) = 60; x 2 - 120 a? + 30 ; 

therefore 1,3, and are the critical values. 

Since /"(1) = -10, /(1)[=0] is a maximum. 
Since' /"(3) = + 90, /(3)[=-28] is a minimum. 
Since /"(0) = 0, and /"'(0) = 30, 

/(0) is neither a maximum nor a minimum. 

5. Examine (a? — l) 4 (a? -f- 2) 3 for maxima and minima. 
Here f(x) = (x - l) 3 (x + 2) 2 (7 x + 5) , 

and the critical values are — 2, — -f-, and + 1. 

In this example, the first method is to be preferred. By in- 
spection, we see that 

f'(—2 — h) and /'(-— 2 + /i) are both positive ; 

hence /( — 2) is neither a maximum nor a minimum (§ 112) c 

f >(-5- h ) is+ , an i /'{_$ + A) is _ ; 

hence /(— "f) * s a maximum (§ 112). 

/'(1-A) is -, and f'(l+h) is +; 
hence /(I) [=0] is a minimum. 

6. Examine & + c(x — a)i for maxima and minima. 

Here /'(*)= , 2 ° ,, ; 

3 (a? — a) s 

and the critical value is a, the root of f'(x) = oo. 

2c . -i 2c . . 

and — is -f ; 



3 (a — h— a)s 3(a+/i— a) 

.*./(«)[= 6] is a minimum. 

(a — x) 3 
7. Examine -^- — - — '— for maxima and minima. 
a — 2 x 

tt ^/ \ (a — x) 2 (4:X — a) 

Here /' (a?) = A > v > ; 

(a — 2 a-) 2 



EXAMPLES. 117 



and the equations fix) = and /'(as) = oo give -, -, and a as 

4 2 
the critical values. 

By inspection, we see that f'(x) changes from negative to 

positive when x = - ; hence ft - ) is a minimum. But, as f\x) 

ct f ol\ 

does not change its sign when x — a or -, f(a) and /f-] are 

neither maxima nor minima. 



8. Examine c -\- V4a¥ — 2aas 3 for maxima and minima. 
By (d), (<?), and (a) of § 114, any value of x that renders 
+ V4 a 2 ;i*- — 2 aas 3 a maximum or a minimum, renders 



V4 aV — 2 aas 3 , 4 eras 2 — 2 aa: 3 , and 2 aas 2 — as 3 

a maximum or a minimum. 

Hence, let f(x) = 2 aos 2 — as 3 , etc. 

-4ns. When x = 0, c+ v / 4a 2 as 2 — 2 aas 3 is a minimum ; 
" cc=|a, u " is a maximum. 

9. What values of as render 2 as 3 — 21 as 2 -f- 36 x — 20 a maxi- 
mum or a minimum? 

Ans. /(I) is a max. ; /(6) is a min. 

10. Examine 3 as 5 — 125 a; 3 + 2 160 as for maxima and minima. 
Ans. /(— 4) and/(3) are max. ; /(— 3) and/(4) are min. 

1 1 . Examine X s — Sx 2 -\-Qx -f-7 for maxima and minima. 

Ans. It has neither a max. nor a min. 

12. If /'(as) = as 3 (as-l) 2 (as-2) 3 (as-3) 4 , what values of x 
render /(as) a maximum or a minimum? 

Ans. /(0) is a max. ; /(2) is a min. 

13. Examine x(x -+- a) 2 (a — x) 3 for maxima and minima. 
Ans. /(—a) and/(- ) are max. ; /( ) is a min. 



118 MAXIMA AND MINIMA. 

14. Examine -2— for maxima and minima. 

aj-10 

Ans. /(4) is a max. ; /(16) is a min. 

(x 4- 2) s 

15. Examine v y n for maxima and minima. 

Ans. /(3) is a max. ; /(13) is a min. 

16. Prove that sin x + cos x is a maximum when x = —- 

4 

17. Examine for maxima and minima. 



log a? 



-Ans. : is a min. 



lose 
18. Prove that of is a maximum when x = e. 



19. Prove that sin #(1+ cos x) is a maximum when x = -- 

o 

X 

20. Prove that is a maximum when x = cos x. 

1 + x tan x 

21. Examine the curve y = x s — Sx 2 — 24 cc + 85 for maxima 
and minima ordinates. 

^Lns. 113 is a max. ; 5 is a min. 

22. Examine y = x 3 — 9 x 2 + 24a? -f 16 for maxima and min- 
ima ordinates. 

.Ans. 36 is a max. ; 32 is a min. 

23 . Examine y = x s — 3 x 2 — 9 x -}- 5 for maxima and minima 

ordinates. 

A.ns. 10 is a max. ; — 22 is a min. 

24. Examine y = x 5 — 5jc 4 + osc 3 + 1 for maxima and minima* 

ordinates. 

Ans. 2 is a max. ; —26 is a min. 

25. Examine y= sin 3 a? cos a? for maxima and minima ordi- 
nates. 

Ans. When x = -J 7?, y = T 3 F V3, a max. 



GEOMETRIC PROBLEMS. 



119 



Geometric Problems. 

1. Find the altitude of the maximum cylinder that can be 
inscribed in a given right cone. 

Let ik be the cylinder inscribed in the d 

given cone dab. Let « = dc, 6 = ac, y = MC, 
x = im, and V= the volume of the cylinder ; 
then V— Trxy 2 . 

From the similar triangles adc and idh, 
we find 

b 



y 



(a - x) 



h- 




r.V=^x(a-xy, 



Fig. 28. 




Fig. 29. 



which is the function whose maximum is required. 

Let f(x) = x (a - x) 2 , etc. § 114, (a) . 

Ans. The altitude of the cylinder = ^ that of the cone. 

2. Find the altitude of the maximum cone that can be 
inscribed in a sphere whose radius is r. 

Let acd and acb be the semicircle and 
the triangle which generate the sphere and 
the cone. Let x = ab, y = bc, and V= 
the volume of the cone ; - then V= \irxy 2 . 

Since y 2 = ab -bd = x(2r— x), 

V=i7rx 2 (2r-x), 

which is the function whose maximum is required. 

Ans. The altitude of the cone = |- the radius of the sphere. 

3. Find the altitude of the maximum c^ylinder that can be 
inscribed in a sphere whose radius is r. 

Let x = ab, and y — be ; 

then V= 2 ivxy 2 = 2 irx(f - x 2 ) . 

Ans. Altitude = f rV3. Fig. so. 



120 MAXIMA AND MINIMA. 

4. Find the maximum rectangle that can be inscribed in ah 
ellipse whose semi-axes are a and b. 

Ans. The sides are <xV2 and &V2 ; the area = 2ab. 

5. Find the maximum cylinder that can be inscribed in an 
oblate spheroid whose semi-axes are a and b. 

Ans. The radius of the base = ^a,-V6-', the altitude == J&V3, 

6. The capacity of a closed cylindrical vessel being c, a con- 
stant, what is the ratio of its altitude to the diameter of its 
base, when its entire inner surface is a minimum? What is its 
altitude ? 

Let y equal the radius of the base, x the altitude, and S the 
entire inner surface ; then 

c = 7rxy 2 , (1) 

and S = 27rif + 2iryx. (2) 

From (1), 

^ = _JL (3) 

dx 2x v ' 

From (2), 

^ = 4^ + 2^3 + 2^. (4) 

dx dx dx 

d S 
Since — = when S is a minimum, from (4) we have 
dx 

dy = y 

dx 2 y -f- x 

From (3) and (5), 

_V_ = -V orx = 2y. (6) 

2x 2y + x v 

Hence, as S evidently has a minimum value, it is a minimum 
when the altitude of the cylinder is equal to the diameter of its 
base. 

From (1) and (6), 



(5) 



\2tt 



GEOMETRIC PROBLEMS. 121 

This problem might have been solved like those preceding it ; 
that is, by eliminating y between (1) and (2) at first. In many 
problems, however, the method given in this example is much 
to be preferred. 

7. The capacity of a c} T lindrical vessel with open top being 
constant, what is the ratio of its altitude to the radius of its 
base when its inner surface is a minimum? 

Arts. Its altitude = the radius of its base. 

8. A square piece of sheet lead has a square cut out at each 
corner ; find the side of the square cut out when the remainder 
of the sheet will form a vessel of maximum capacit}'. 

Ans. A side = -J- the edge of the sheet of lead. 

9 . Find the arc of the sector that must be cut from a circular 
piece of paper, that the remaining sector may form the convex 
surface of a cone of maximum volume, r being the radius of the 

circle ' Ans. The arc = 2 ttv (1 - JV6) . 

10. A person, beiug in a boat 3 miles from the nearest point 
of the beach, wishes to reach in the shortest time a place 5 miles 
from that point along the shore ; supposing he can walk 5 miles 
an hour, but row only at the rate of 4 miles an hour, required 
the place where he must land. 

Ans. 1 mile from the place to be reached. 

11. Find the maximum right cone that can be inscribed in a 
given right cone, the vertex of the required coue being at the 
centre of the base of the given cone. 

Ans. The ratio of their altitudes is -J-. 

12. A Norman window consists of a rectangle surmounted by 
a semicircle. Given the perimeter, required the height and the 
breadth of the window when the quantit} r of light admitted is a 
maximum. 

Ans. The radius of the semicircle = the height of the rectangle. 



122 



MAXIMA AND MINIMA. 



13. Prove that, of all circular sectors having the same 
perimeter, the sector of maximum area is that in which the cir- 
cular arc is double the radius. 

14. Find the maximum convex surface of a cylinder inscribed 
in a cone whose altitude is 6, and the radius of whose base is a. 

Ans. Maximum surface = ^irab. 

15. Find the altitude of the cylinder of maximum convex 
surface that can be inscribed in a given sphere whose radius 
is r. Ans. Altitude = rV2. 

16. Find the altitude of the cone of maximum convex surface 
that can be inscribed in a given sphere whose radius is r. 

Ans. Altitude = |r. 

17. Find the altitude of the parabola of maximum area that 

can be cut from a given right circular 
cone. 

Let ob = 2 6, oc = a, and bm = x ; then 

qq' = 2 mq = 2 Vmb • MO 




= 2^/x(2b-x). 
Also, bo : bm : : oc : mp, 
2 b : x : : a : mp ; 



.*. MP = 



2a 



ax 
Yb 



.'. area = f qq'- mp = — -yx^{2b — x). 

O 



§ 66, Ex. 1. 
§ 114. 



Let f(x) = x s (2 b — x) , etc. 

Ans. The parabola is a maximum when its altitude mp is f 
the slant height of the cone. 

18. What is the altitude of the maximum cylinder that can 
be inscribed in a given prolate spheroid ; that is, in a solid gen- 
erated by the revolution of a given ellipse about its major axis ? 
Ans. Altitude = the major axis divided by V3. 



GEOMETRIC PROBLEMS. 123 

19. Find the number of equal parts into which a given num- 
ber a must be divided that their continued product may be a 

maximum. , „. , _ , a 

Ans. I he number of parts = -, and each part = e. 

20. A privateer has to pass between two lights a and b, on 
opposite headlands. The intensity of each light is known, and 
also the distance between them. At what point must the pri- 
vateer cross the line joining the lights, so as to be in the light 
as little as possible ? 

Let d = the distance ab, and x the distance from a of any 
point p on ab. Let a and b be the intensities of the lights a 
and b respectively, at a unit's distance. By a principle of 
Optics, the intensity of a light at any point equals its intensity 
at a unit's distance divided by the square of the distance of the 
point from the light. 

Hence 1 is the function whose minimum we seek. 

x 2 (d — xy 

Ans. x= dai ■ 
as + U 

21. The flame of a lamp is directly over the centre of a cir- 
cle whose radius is r ; what is the distance of the flame above 
the centre when the circumference is illuminated as much as 
possible ? A 

Let a be the flame, p any point on the circum- 
ference, and aj = AC. By a principle of Optics, 
the intensity of illumination at p varies directly 
as sincPA, and inversely as the square of pa. 

Hence — is the function whose maxi- 

( ? " + ar ) J . . Fig. 32. 

mum is required, in which r is the radius of the 

circle, and a is the intensity of illumination at a unit's distance 

from the flame. An ^ x= i r ^. 

22. On the line joining the centres of two spheres, find 
the point from which the maximum of spherical surface is 
visible. 




124 



MAXIMA AND MINIMA. 



Let cp = r, cp = _R, cc = d, and ca = x, a being any point on 

mM. From a draw the tangents Ap and ap ; then the sum of the 

zones whose altitudes are nm 

and nm is the function whose 

maximum is required. 

r 2 
Since cn = — , by Geometry 

x 

we have 




Hence 2-n- 



zone nm = 2-kt • nm = 2irr(7 



en) = 2 7r ( r 2 — 



r 2 +R 2 



imum is sought. 



d — x 



is the function whose max- 
dri 



Ans. x 



+ i?i 



23. Assuming that the work of driving a steamer through 
the water varies as the cube of her speed, find her most eco- 
nomical rate per hour against a current running c miles per 
hour. 

Let v — the speed of the steamer in miles per hour. 

Then av s = the work per hour, a being constant ; 
and v — c = the actual distance advanced per hour. 

av s 



Hence — = the work per mile of actual advance. 

v — c 



Ans. r o — \c. 



CHAPTER IX. 

FUNCTIONS OF TWO OR MORE VARIABLES, AND CHANGE OF 
THE INDEPENDENT VARIABLE. 

115. Functions of two or more Variables. Since any inde- 
pendent variable is some arbitrary function of t, t representing 
time, a function of an}' number of independent variables may 
be regarded as a function of the single variable £, and therefore 
differentiated by the rules already established. 

/(a?, ?/), read "function of x and y" represents any function 
of x and y\ as, x 3 -\- xy 2 -\- xy and sin(ic -f- y) . 
/(ic, y, z), read "function of jc, y, and z," represents any 
and z. 

116. A Partial Differential of a function of two or more vari- 
ables is the differential obtained on the Irypothesis that only one 
of the variables changes. 

117. A Total Differential of a function of two or more vari- 
ables is the differential obtained on the hypothesis that all its 
variables change. 

118. A Partial Derivative of a function of two or more vari- 
ables is the ratio of the partial differential of the function to 
the differential of the variable that is supposed to change. 

119. A Total Derivative of a function of two or more vari- 
ables, of which only one is independent, is the ratio of the total 
differential of the function to the differential of the independent 
variable. 

If u = f(x, y), the partial differentials of u with respect to 

x and y are written d x u and d it, or — dx and — dy; and the 

dx dy 



126 FUNCTIONS OF TWO OR MORE VARIABLES. 

partial derivatives, or differential coefficients, are written 

d x u -, d v it . A du -. du 

- JL - and - 1 -, or simply — and — 
dx dy " dx dy 

120. The total differential of a function of two or more vari- 
ables is equal to the sum of its partial differentials. 

For, if u =/(#, y) , it is evident, from the general principles of 
differentiation, that du can contain only such terms as are of the 
first degree in dx and dy. 

Hence du = <£(#, y)dx -f- <f>i(x, y)dy, (1) 

in which <£(#, y) and <f>i(x, y) represent the sums of the co- 
efficients of dx and dy in the different terms of du. 

Let x' and y' be any set of values of x and y ; then we have 

du = <f>{x\ y')dx -f faQc'-, y')dy. (2) 

Let y be regarded as constant ; then dy = 0, 4>(x', y') remains 
unchanged, and (2) becomes 

d x u = <f>(V, y')dx. (3) 

If x is constant, (2) becomes 

d y u=cf, 1 (x\ y')dy. (4) 

Adding (3) and (4), and remembering that x' and y f are any 
values of x and y, we have in general 

d x u -f- d y u = </>(«, y)dx + <f>i(x, y)dy = du. 

Since a similar process of reasoning could be extended to a 
function of n variables, the theorem is proved. 

If x and y were not independent, the demonstration given 
above would still hold ; for the idea of a partial differential of a 
function sets aside any question concerning the dependence of 
its variables. 

121. Signification of Partial Derivatives. From § 31 it is 
evident that a partial derivative expresses the ratio of the rate 



THE TOTAL DERIVATIVE. 127 

of change of the function to that of its variable, so far as its 
rate depends on the variable supposed to change ; and that a 
function is an increasing or a decreasing function of any one of 
its variables, according as its partial derivative with respect to 
that variable is positive or negative. 

Examples. 

1 . u = by 2 x -j- ex 2 -f- gy 3 -\- ex ; find du. 
Here d x u = {by 2 + 2 ex -J- e) dx, 

and d y u = (2 byx -f 3 a?/ 2 ) cfa/ ; 

.-. dw = {by 2 + 2cx-j-e)dx-\-{'2 byx -f 3 gy 2 ) dy. 

2. u = y x . Ans. du = y x log ydx-\- xy x ~ x dy. 

y 

3. ^ = logcc 2 '. dw = -d#-f-loga?cfa/. 

v x dy — y dx 

4. w = tan~ 1 -- du = , , 9 • 

aj ar + 1/ 2 



5. u = y slnx . du = y smx logy cos xdx-\ dy. 

° iJ tsiJ * overs* •* 

: ,aj ydx — xdy 

6. it = log tan 1 -. aw = — 



(a^ 2 + 2/ 2 )tan- 1 ^ 



7. W = _4-J-. d^ = — cfa + -fay 

a 2 6 2 a 2 6 2 

122. If u =f{x, y, z). and y=<j){x), and 2 = <£!(#), it is 
directly a function of aj, and indirectly a function of a? through 
y and z. If u=f{z, y), and y = <j>(x), and 2! = <^> 1 (ic), w is, 
in like manner, indirectly a function of x through y and z. In 
all such cases the total derivative of u with respect to x can be 
obtained by finding the value of u in terms of x, and differen- 
tiating the result ; but in man}' cases it is more readily obtained 
by using the formulas of the next article. 



128 FUNCTIONS OF TWO OR MORE VARIABLES. 

123. If u=-f(x i y, z), and y = <f>(x), and z = <f> 1 (x), 

i j du -, du -, . du , 

we have du = — dx-\ dy -\ dz ; §120 

dy dz 

du dy du dz 



Vdu~\ *_ du 
' [_dxj dx 



+ 



dy dx dz dx' 



in which 



du du 



du 



dx 



— , and — - are the partial derivatives, and 
dy dz 



the total derivative of u. 



du" 
dx 



Cor. 1. If u=f(x,y), and y = <f>(x), du = —dx + —dy: 

dx dy 



Fdu~\ __du du dy 
\_dxj dx dy dx 



Cor. 2. If u=f(y, z), and y=<f>(x), and z = <f> 1 (x), 

■, du -, . du -j 

du = — dy -\ dz ; 

dy dz 



du 

dx 



_ du dy du dz 

dy dx dz dx 



Cor. 3. If u = f(y), and y=<j>(x), du 

du _ du dy 
dx dy dx 



du 
dy 



dy; 



Rem. To make the above theorem and corollaries intelligible, 
the signification of each term and factor must be had clearly in 

*~diT 



mind. Thus, in the theorem, 



dx 



denotes the total derivative 



of u with respect to a?, that is, the derivative obtained on the 

hypothesis that x, y, and z are all changing according to the 

du 
given conditions ; while — denotes the partial derivative of u 

with respect to a?, that is, the derivative obtained on the hypo- 
thesis that y and z are constant. 



* Analysts are not agreed as to the best means of distinguishing total 
from partial derivatives ; but we shall always distinguish the total deriva- 
tive of a function of two or more variables by enclosing it in brackets. 



EXAMPLES. 129 



Examples. 

1. u = z 2 -r y s + zy, z = sinx, and y = e x ; find 

TT \~du~] dudy . dicdz /iN 

Here — = *H (1) 

\_dx_] dy dx dz dx 

— = Sy 2 + z, — = 2z + y, 
dy dz 

dz , dii , 

— = cosoj, and -*-=&'. 
dx dx 

Substituting these values in (1), we have 

[^ = (3y* + z)e* + (2z + y)cos'x 

= (3e 2x + sin^)e x + (2sinx + e a: )cosa? 
= Se 3x -\-e x (sinx-{-cosx) -+- sin 2 x. 

This same result could be obtained by substituting in u the 
values of y and z in terms of a?, and then differentiating. 

2. u = tan" 1 (xy) , and 2/ = e x . ^4ws. f— 1 = g-Ji±g) . 

(_(iccj l-j-ar^e 2 * 

3. it = e ax (y — z), y = a since, and2=cos#. 

r^~| = (a 2 + l)e-sin^ 



4. % = tan _, and # 2 4- if = ?- 2 . 



r^i=-ior =1 

Ldaj J y V?- 2 - 



5. w = sin-, z = e x , and y = x 2 . 



6. u = -yx 2 -f- y 2 , and y = mx + c. 

mc_ 



— = (x— 2) — cos—-! 
dx] V J x* a; 2 

f^!f~l _ (1 + m 2 )x + mc 
L^ J Var -j- (ma? -f- c) : 



130 



FUNCTIONS OF TWO OR MOKE VARIABLES. 



7. w = sin 1 (y — z), y=3x, and z 



a i fdu 
= 4x 3 . — 
\_clx_ 



VI -x 2 



x"y 



u = a;*?/ 2 ^- + x 4 , and y = logx. 

d?£ 
dx 



ar 1 [4 (log a;) 2 + 3£]. 



124. Implicit Functions. The equation /(oj, y) = indicates 
that either y or x is an implicit function of the other. It 
represents an}' equation containing x and y when all its terms 
are in the first member. Implicit functions have been differen- 
tiated heretofore, but the formula in § 125 is often useful in 
obtaining the first derivative of an} 7 such function. 



125. If u=/(a>, ?/), du 



du 7 , du 7 

— dy H dx. 

dy " dx 



(i) 



If u is constant, du = 0, and from (1) we obtain 

du 

dy = _dx 

dx~ du 

dy 

in which — and — are partial derivatives. 
dx dy 

When u is constant, du = d x u -f- d y w = ; but, in general, 
neither d x ii nor d y u is zero. 



Examples. 



1. y 3 -2x 2 y + bx = 0; find ^- 

Here ^=-4^ + 6, ^ = 3 2 / 2 -2aj 2 ; 
dx d?/ 

dw- 



dy _ dx _ 4yx—b 
dx~~du = 3y 2 -2x 2 ' 
dy 



IMPLICIT FUNCTIONS. 131 

2. X s + y 3 — 3 axy = c — u. 






4. # log y — 2/ logo; = 0. 

5. f(ax+by) = c. 

Here f(ax -f- 6?/) = c = u; 





dy _x 2 - 


-ay 




dx ax 


-2/ 2 


dy 

dx 


--en 


ST 


dy _ 


2/ /# log ?/ - 


-.v\ 


dx 


a%loga- 


-J 





' ' dx 


/'((MJ 


+ ty) 


a 




dy _ 
t7x 


a 
6* 






6. 


a* — 2/* = 








7. 


a 3 + 3 axy = 


= -2/ 3 







-^=f'(ax + by)b; 



d v = y 2 — qyiogy , 

da; x 2 — xy log a; 

dy _ cc 2 + ay 
d"# y 2 4- aa; 

126. General formulas for the successive derivatives of an 
implicit function can be easily deduced ; but they are so com- 
plicated that, in practice, it is generally more convenient to 
differentiate directly the first derivative to obtain the second, 
and so on, than to use these formulas. 

Examples. 

d 2 v 
1 . Find — - when y 2 — 2 xy -j- c = 0. 
dx 2 

Here §* = -Z- (1) 

dx y — x 



Differentiating (1), we have 

(y — x) ~y~ — y ( -*- — 1 ) y — x — *- 
d 2 y _ dx y \dx J _ _ dx 

dx 2 (y—x) 2 (y-x)~ 



(2) 



132 



FUNCTIONS OF TWO OR MORE VARIABLES. 



From (1), (2), and the given equation, we obtain 

d 2 y _ y 2 — 2 xy _ — c 
dtf~\y-xy~ (y-xy 

2. Given y 2 — 2 axy + x 2 — c = 0, to find 



a^/ # 
dx 2 



d 2 y _ (a 2 - 1) (y 2 - 2 gay + ^ 2 ) = c (a 2 - 1) 
dx 2 (y — ctx ) 3 ( 7/ — ax) 3 



3. Given y 3 + a? — 3 ax?/ = ; show that -~ = — , 2 a ^ x , 
* eta 2 (y 2 — axy 



127. Successive Partial Differentials and Derivatives. If 
u—f(x,y), x and y being independent, d B w and d y ii are, in 
general, functions of both x and ?/. Differentiating d x u and d y w 
with respect to either variable, we obtain a class of second par- 
tial differentials. Ity differentiating these second partial differ- 
entials, we obtain a class of third partial differentials ; and so 
on. In finding these successive partial differentials, we regard 
dx and dy as constant, since we may suppose x and y to change 
uniformly. 

The successive partial differentials are represented as follows : 

d 2 u 



7 I'du 7 

d x [ — dx 

dx 



, ,'du 7 



, fd 2 u 7 . 

dj — dy 

\dy 2 

A—dy 2 



dy- 



d fdu 7 \ 7 " 
= — — ax) dx 
dx \dx ) 

= AM d x)dy 

dy\dx J 

d fd 2 u 7 2 \ 7 



^A(^dy 2 )dy 
dy\dy 2 U ) \ 



dx 2 



dx 2 ; 



d 2 u 
dxdy 

d 5 u 



dy'-dx 
dhc 



dx dy ; 
dy 2 dx ; 



dif 



dy 3 ; etc. 



Hence —dxdy 2 is a symbol for the result obtained by dif- 

dxdy 2 

ferentiating u—f(x,y) three times in succession: first, once 
with respect to x, and then twice with respect to y. 



PARTIAL DIFFERENTIALS AND DERIVATIVES. 133 

The symbols for the partial derivatives are 
dru d 2 u d 2 u dht d s u 



dor'' dxdy' dy 2 ' dar 3 ' dx*dy 



etc. 



1 OR Tf — f ( * \ ^' U - - ^ U < ^ U ^ U t 

dydx dxdy' dydx 2 dardy' 

that is, if u be differentiated m times ivitJi respect to x, and n 
times with respect to y, the result is the same, ivhatever be the 
order of the differentiations. 

Y(x + bx,y)-f(x,y) ' 
Ax 



For limit 



dx A * = ° 

Regarding this expression as a function of y, finding its incre- 
ment, dividing by Ay, and remembering that the difference 
between the limits of two variables is equal to the limit of their 
difference, we have 

d_(du y 
dy\dxj 

f limit [ 71*+**, y+Ay)-f(x, y +A y)-f(x+Ax, y)+f(x, y ) " \ 
limit \ Ax = o[_ Ax J I 



A3/-0^ A y j 

In like manner, we should obtain for the value of — f — ) the 

dx\dyj 

same expression, except that the order of the limits would be 
reversed ; but, from the nature of the process of passing to a 
limit, it is evident that this change of order does not affect the 
value of the expression. 

TT d fdu\ d fdu\ d 2 u d 2 u , n x 

Hence — — = — f — ] , or = ( 1 ) 

dy\dxj dx\dyj dxdy dydx 

Again, since 

d 2 u dru 



dxdy dydx" 1 
d 3 u d 3 u 



dxdydx dydx 2 



(2) 



134 FUNCTIONS OF TWO OR MOKE VARIABLES. 

But, from the principle in (1), we have 

d d (du\ _ d d fdu\ d 3 u d 3 u ,„\ 

dx dy\dx) dy dx\dxj^ dxdydx dx 2 dy 

From (2) and (3), 

d 3 u d 3 u 

dy dx 2 dx 2 dy 

This method of reasoning evidently applying to all cases, the 
theorem is established. 



1 . u = cos (x -f- y) ; verify 

2. u = x'y 2 -f- ay 5 ; verify 



Examples. 

d 2 u d 2 u 



dydx dxdy 
d 3 u d 3 u 



dy 2 dx dxdy 2 

3. u = log (x -\- y) ; find the first, second, and third partial 
derivatives. 



4. u = tan l ^- ; verify 



d 3 u d 3 u 



5. u = sin (bx 5 -f ay 5 ) ; verify 



dx 2 dy dy dx 2 

dhi d 3 u 



dyhlx dx dy* 



129. To find the successive differentials of a function of two 
independent variables. 

Let u =/(#, y) ; then 

du = — dx + ^dy. (1) 

dx dy 

du du 

Differentiating (1), remembering that — and — are, in gen- 
eral, functions of both x and y ; and that, as x and y are inde- 
pendent, dx and dy may be regarded as constant, we have 



CHANGE OF THE INDEPENDENT VARIABLE. 135 

d 2 u-,o , dru , , . dru 7 7 . dru , 2 
= — dor -\ dxdy -\ dydx -\ dy- 
dx- dxdy dydx dy 

= ^dx* + 2 £L dxdy + f^<% 2 . (2) 

Differentiating (2), remembering that, in general, each term 
is a function of both x and y ; and that the total differential of 
each is the sum of its partial differentials, we obtain 

$u = —da? + 3 -^dx 2 dy+ 3 -^ dxdy 2 +—dtf. 
dx? dx 2 dy dxdy dy 2. 

Differentiating this equation, we obtain an analogous expres- 
sion for dhi, and so on. 

B} r observing the analogy between the values of d 2 u and d 3 u, 
and the development of the second and third powers of a bino- 
mial, the formula for d n u may be easily written out. 



130. Change of the Independent Variable. The forms of the 

successive derivatives of -^ used thus far have been obtained 
dx 

upon the hypothesis that x is the independent variable, and that 
dx is constant. In applications of the Differential Calculus, it 
is often desirable to change the independent variable, and to 
regard the original function, or some other variable, as the inde- 
pendent variable. We proceed to find the forms of the succes- 

du 
sive derivatives of -£ upon other hypotheses than that dx is 

constant. 

131. To find the successive derivatives of —2 ; that is, the forms 

dx 

of — (— ], — • — f— 1, etc., when neither x nor y is indepen- 
J dxVdx/ dx dxVdx/ 
dent. 

If neither x nor y is independent, -^ is a fraction with a vari- 

dx 

able numerator and denominator ; and we have 

d fdy\ __ dxdry — dydrx , 1 v 

dxldx J dx s ^ j 



136 CHANGE OF THE INDEPENDENT VARIABLE. 

Differentiating (1), we obtain 

d_ _ d_ fdy\ _ _d_ fdxd 2 y — dy d 2 x 
dx dx\dxj dx\ dx" 

_ ( d"y dx — d s x dy) dx—3( d 2 y dx — d 2 x dy ) d 2 x (i> . 

dx 5 { ' } 

In like manner, we obtain the other successive derivatives. 

Cor. 1. If y is independent, that is, if dy is constant, 

d 2 y = 0, and d s y = 0, 

and (1) and (2) become 

d fdy\ = d 2 xdy ( . 

dx\dx) cW ' K } 

i sL - f^v\ — - ( c ^ 2a; ) 2 c ^ ~ ^ x c ty ^ x (±\ 

dx dx\dxj dx 5 

Cor. 2. If dx is constant, d 2 £ = <#U'=0, and (1) and (2) 
become 

A(dy) = S, and — ~( r ^\ = ^V 
dx\dxj dx 2 " 1 dx dx\dx) dx^ 

which agrees with § 74. 

Rem. Hence, to transform a differential expression in which 
x is independent, into its equivalent in which neither x nor y is 

independent, we replace — - , — «, etc., bv their equivalents upon 
dx~ dx 3 

the new hypothesis, which are found in (1), (2), etc. 

If, in the transformed expression, a new variable #, of which 
x is a function, is to be the independent variable, in the general 
result obtained above we replace x, dx, d 2 x, etc. , by their values 
in terms of and its differentials. 

If y is to be the independent variable, in the given expression 

dry d?v 
we replace — ~, —^ etc., by their equivalents in (3) , (4) , etc. ; or, 

CIX" CIX 

in the general result first obtained above, we put d 2 y = 0, d s y = 0, 
etc. 



EXAMPLES. 137 



Examples. 



1. Given yd 2 y -f dy 2 + dx 2 =0, in which x is independent, to 
find the transformed equation in which neither x nor y is inde- 
pendent ; also the one in which y is independent. 

Dividing both members by dx 2 , substituting for — * the sec- 
B J to dx 2 

ond member of (1), § 131, and multiplying both members by 
dx 3 , we have 

y(d 2 ydx — d 2 xdy) + dy 2 dx-\- dx 3 = 0, 
in which neither x nor y is independent. 



Putting d 2 y = 0, and dividing by — dy 3 , we have 
dy 2 dy 3 dy 



in which the position of dy indicates that y is independent. 



2. Given — ^ -^ + — ^— =0, in which x is indepen- 

dxr 1 —x? dx 1 — x 2 

dent, to find the transformed equation when x = cos 0, and is 

independent. 

dry 
Substituting for — % the second member of (1), we have 
dx 2 

d 2 ydx — d 2 xdy x dy V _ q /-i\ 

' dx 3 1 - x 2 dx 1 - x 2 ~ ' ^ j 

Since a; = cos 0, 1 — # 2 = sin 2 0, 

cZic = — sin dO, and fe = — cos d6 2 . 

Substituting these values in (1), and simplifying, we have 



in which 6 is independent. 



138 CHANGE OE THE INDEPENDENT VARIABLE. 



3. Given B = 



1 + df 
dx 2 



in which x is independent, to find 



dry 

dx 2 
the value of R when .T = pcos0, 2/ = psin#, and 6 is inde- 
pendent. 

From (1), § 131, 

R = {dx> + dy 2 )% ^ _ 

dx d 2 y — dy d 2 x' 
in which neither x nor y is independent. 
From y = p sin#, and x = p cos 0, we obtain 
cfa/ == sin 6dp-\- p cos d#, 
dx = cos Odp — ps'mO dO, 
d 2 y = sin 0d 2 p + 2 cos (9 d<9 dp - p sin 6 dO 2 , 
and d 2 # = cos d 2 p — 2 sin dO dp — p cos d# 2 . 

Substituting these values in (1), and simplifying, we have 



+p 2 



dl 
dO 2 



p+ d0 2 P d0> 

4. Given — %-\ ^-f2/ = 0, to find the transformed equa- 

dx 2 xdx 

tion when x 2 — 42, and z is independent. 

dz- dz 

5. Given (1 - x 2 ) ^ - x^- = , to find the transformed 

dxr dx 

equation when x = cos 2, and 2 is independent. 

6. Given 2 = ' — " ~^ — , to find the transformed equation 

ydy -f- #dsc 
when # = p cos 0, y = p sin 0, and p is independent. 

Ans. z = *f- 
dp 



CHAPTER X. 
TANGENTS, NORMALS, AND ASYMPTOTES. 

132. The Rectangular Equation of the Tangent to any plane 
curve at (x\ y') is 

dv'* 

y-y'=-^(v-x')' (a) 

For line (a) passes through {x\ y') ; and, by § 16, it has the 
slope of the curve at (V, y'). 

dv' 
Cor. When the axes are oblique, — is evidently the ratio 

dx' 

of the sines of the angles which the curve at (V, y') makes with 
the axes ; hence, in this case also, («) is the equation of the 
tangent. 

133. The Rectangular Equation of the Normal to any plane 
curve at (V, y') is 

y-y'---^-,(x-x'). (b) 

For, the axes being rectangular, line (b) is perpendicular to 
line (a) of § 132, and therefore to the curve, at (x 1 , y'). 

Examples. 

1. Find the equations of the tangent and normal to the 
parabola y 2 = 2px. 

Here *L=l; .:*>!-? 

dx y dx' y' 



— represents the value of c -^ at the point (x' } y'). 
dx' dx 



140 TANGENTS, NORMALS, AND ASYMPTOTES. 

civ' 
This value of — ( substituted in (a) of § 132, and (b) of 
dx' 

§ 133, gives 

jf-y'= J (*■-«'). (i) 

y' 

and y-y'=--(x-x'), 

as the equations of the tangent and normal respectively. 
Since y' 2 = 2px', equation (1) by reduction becomes 
yy'=p{x + x'). 

2. Find the equations of the tangent and normal to the circle 

9 i 9 9 

xr -\-y = r. 

v r 

Ans. yy' -{- xx f = ^ ; y = — f x. 
x 

3. Find the equations of the tangent and normal to the 
ellipse a 2 y 2 + 6 2 ar = orb 2 . 

Ans. a 2 yy'-\- b 2 xx' = orb 2 ; y — y'= —f- (x — x') . 

4. Find the equations of the tangent and normal to the 
hyperbola a 2 y 2 — b 2 a? = — a 2 b 2 . 

Ans. a 2 yy' — b 2 xx' = — a 2 b 2 ; y — y'= — -f- (x — x') . 

b~x' 

5. Find the equations of the tangent and normal to the cis- 
soid y 2 — 



2a — x A , .x'zCSa — x'), , N 

Ans. y — y' = ± — * — '- (x — x') ; 

J J (2a-x')i V ; 

, (2 a — x') i , , x 

x'*(3a — x') 

6. Find the equation of the tangent to y 2 = 2 x 2 — x 2, at x = 1. 

Ans. y = ix + i', y = — \x — \. 

7. Find the equation of the normal to y 2 — 6 x — 5 at y = 5. 

Ans. y = — f»H-^. 



LENGTH OF SUBTANGENT AND SUBNORMAL. 141 

8. Find the equation of the tangent to the hyperbola referred 
to its asymptotes, xy = m. ^ y = _^ x+2y ,_ 

X 

9. Find the equation of the tangent to the cycloid 



x = '.* vers -1 — — V 2 ry — y 2 . Ans. y — ?/'= A/ j-^- (x — x') . 



134. Length of Subtangent, Subnormal, Tangent, and Normal 
Let pt be the tangent at the point p 
(x',y'), and ps the normal. Draw the 
ordinate pm ; then tm is called the sub- 
tangent, and ms the subnormal. 

mp , dx f 

tm = = y — - ; 

tan mtp dy' 




dx' 
subtangent = y' — -• (I) 

dy' v ' 

ms = mp tan mps = y' tan mtp = y' -&- ; 
* J dx' ? 

subnormal = y'—' (2) 

dx' v 7 



PT = V MP"' -f- TM 2 = Ay' 2 -j- ( l 



,dxy 

' dy'J ; 
•. tangent = ^l+^;j- (3) 

ps = V^p 2 + ms 2 =-\jy' 2 + (V j-A ; 

.-.normal = 2/^1 + ^. (4) 

Rem. If the subtangent be reckoned from the point t 
(Fig. 34), and the subnormal from m, each will be positive or 
negative according as it extends to the right or left. 



142 TANGENTS, NORMALS, AND ASYMPTOTES. 

Examples. 

1. Find the values of the subtangent and subnormal of the 
ellipse a 2 y 2 + b 2 ar = orb 2 . 

c , , , dx' cry' 2 x' 2 — a 2 

Subt - =?/ ^ = -^v = ^r- ; 

o x. i dif b 2 x' 

2. Find the values of the subtangent and subnormal of the 
parabola, circle, hyperbola, and cissoid. 

Ans. Parabola : subt. = 2x' ; subn. =p. 

Circle : subt. = — ^— ; subn. = — x'. 

x' 

»'2 2 7,2 I 

Hyperbola : subt. = — ; subn. = 

x' or 

n . ., ,, x'(2a-x') , x' 2 (3a-x') 

Cissoid : subt. = — * - ; subn. = — * '- . 

3 a -x[ (2a-x') 2 

3. Find the value of the subtangent of the logarithmic curve 
y = a x ; also of y 2 = Sx 2 — 12 at a; = 4. ^ns. m; 3< 

4. Find the values of the subnormal and normal of the cycloid. 



Subnormal = V (2 r — y)y = Vhb • hd = ph = ed, 



Normal = pd= v ed + ep" = a/2 ry. 

Thus the normal passes through the foot of the vertical diam- 
eter of the generating circle, when it is in position for the 

point to which the normal 
T , „ /^T*^ ~~~~-\ * s drawn. Moreover, since 

dpb is a right angle, the 
tangent passes through the 
other extremity of the ver- 
, tical diameter. This prop- 
erty furnishes a simple 
method of drawing a tangent and normal to the cycloid at any 




POLAR CURVES. 



143 



point. Thus, to draw a tangent and normal at p, put the gen- 
erating circle in position for this point, and draw the vertical 
diameter bd. The lines drawn from b and d through p will 
be respectively the required tangent and normal. 

That pb is tangent to the cycloid at p is further evident; 
since, when the generating point reaches the position p, it is 
rotating about the point d, and is therefore moving in a direc- 
tion perpendicular to dp. 

135. Fundamental Principle in the Method of Limits. Let 

a, a n /5 and /5 15 be any four variables, so related that 

limit — = 1 , limit -=- = 1 , and limit -5 = c : 
a a ay fii a x a fi 1 

limit- = limit -^ X limit— x limit^ = limit-^-- 
P . Pi - *i P Pi 



then, since 



Hence, in any problem concerning the limit of the ratio of 
two variables, either may be replaced by any other variable, the 
limit ofivhose ratio to it is unity. 



136. Length of Subtangent, Subnormal, Tangent, and Normal 
in Polar Curves. Let ox be 
the polar axis, and p an}' point 
on the curve mn. Let arc pd 



= As, and ob 



then arc 



be = A0, arc pm = pA0, and 
md = Ap. Draw the chords pm 
and pd, the tangent pz, and 
ph perpendicular, and zh par- 
allel, to op, thus forming the 
right triangle pzh ; then 

P A0 



limit 
A0 = O 



chord PM 



=1, §48. 




Fig. 36* 



144 



TANGENTS, NORMALS, AND ASYMPTOTES. 



and 



and 



limit 


As 


— ] 


As = 


chord pd 




limit 

A0=O 


>A<9~ 

_ As_ 


, limit 

~~ A6 = 


limit 
A9 = 


rAp- 

As 




limit 
~ A9 = 





chord pm" 
chord pd 



§ 135. 



MD 



chord pd 



} 



The limit of angle mpd is evidently hpz ; and, in the isosceles 
triangle pom, the limit of angle pom being zero, the limit of 
pmo or its supplement pmd is a right angle. Hence the angles 
of the triangle hpz are the limits of the angles of the triangle 
mpd. Therefore the ratios of the sides of the triangle hpz 
equal the limits of the ratios of />A0, As, and A,o. Hence these 
sides may be taken as pdO, ds, and dp. 

Draw ot perpendicular to op, and produce it until it meets 
the tangent in t. Draw also the normal pa, and the perpen- 
dicular on upon the tangent. The lengths pt and pa are called 
respectively the polar tangent and polar normal; oa is the polar 
subnormal, and ot the polar sabtangent. 

P d0 



tan opt = tan hzp = — = 

hz dp 



HP 



dO 





sin opt = sin hzp 


ZP 


ds ' 




From 


triangle 
ds 2 = 


HPZ, 

dp 2 + P W. 










0T = 


op tan opt = 


2 d0 
dp 








.*. polai 


subtangent 


2 d0 

= p" — 

H dp 








OA = 


op tan opa = 


- OP COt OPT = 


dp . 
dO J 




.*. polai 

PT = 


subnormal = 


_ dp. 

= do' 








V OP + OT = 


-Jp 2 + 


,d0 2 

PT-9 


j 



(1) 

(2) 
(3) 

(4) 
(5) 



dp 2 



POLAR CURVES. 145 



.*. polar tangent = p-v/l + p 2 -rr (6) 

\ dp- 






.-. polar normal = -Jp 2 + -£■■ (7) 

oup p"Clu /o\ 

« = on = OP Sill OPT = p z — = — r — (o) 

137. That the sides of the triangle pzh (Fig. 36) may be 
taken as ds, dp, and pdO can be proved also as follows : 

When the generatrix of the curve is at p, the radius vector 
is increasing in length in the direction of pr, and the extremity 
of the radius vector drawn to p is moving in the direction of ph, 
at the rates at which the generatrix is then moving in these 
directions. If, at p, the motion of the generatrix became uni- 
form along the tangent pz, it is evident that any simultaneous 
increments of its distances from p and lines ph and or may 
be taken as ds, dp, and the differential of the arc traced by the 
extremity of the radius vector to p, which equals pdO; for, 
if ob = 1, b describes the measuring arc of the variable angle 0, 
and p is moving, as op revolves, p times as fast as b. Hence 
ph = pdO, and hz = dp, if pz = ds. 

Examples. 

1. Find the subtangent, subnormal, tangent, normal, and 7?, 
or the length of the perpendicular from the pole to the tangent, 
of the spiral of Archimedes p = aO. 

subnormal = — = a ; 

dO 

subtangent = p 2 ^ = ^ ; 
dp a 



146 TANGENTS, NORMALS, AND ASYMPTOTES. 



tangent = p^jl + p 2 — = pyjl + ^ ; 

2 

normal = V/o 2 -\-a 2 ', j9 = — ^— 



V p 2 + a 2 



2. Find the subtangent, subnormal, tangent, and normal of 
the logarithmic spiral p = a 9 . 

Ans. subt. = — — ; subn. = p log a ; 

log a 



tan. = pil + - ; nor. = p VI + (log a) 2 . 

\ (loga)^ 

Since tan opt =^ — = , this curve makes the same angle 

dp log a 

with every radius vector, and therefore is called the equiangular 

spiral. 

If a = e, tan opt == 1, opt = -, subtangent = subnormal, and 
tangent = normal. 



3. Find the subtangent, subnormal, and p of the lemniscate 

of Bernouilli p 2 = a 2 cos 20. 

— o s —a 2 

Ans. subt. = H — ; subn. = sin 2 ; 

a 2 sin 2 p 

P 3 P s 



V^-f « 4 sin 2 20 « 



Rectilinear Asymptotes. 

138. A Rectilinear Asymptote is a straight line that has the 
limiting position of the tangent to an infinite branch of a curve. 
If a curve has no infinite branch, it evidently can have no 
asymptote. 

If X and Y represent the intercepts of a tangent on the axes 
of x and y respectively, from the equation of the tangent to 
any plane curve 



ASYMPTOTES. 147 

dx' 

dx' 
we obtain X=x t — y , -—i (1) 

dy' 

and Y=y'-x' c ^. (2) 

dx 

Now if, as the point of contact (V, y') moves out along an 
infinite branch, the value of X or Y or the values of both 
approach finite limits, it is evident that these limits will be the 
intercepts on the axes of an asymptote to that branch of the 
curve. 

Examples. 



1 . Examine y s = 6x? + X s for asymptotes. 
Solving for y, we have 



2/ = *% + 



\x 

As x == co,* y = oo ; and, as#= — oc,?/= — oo. 

Hence the curve has two infinite branches, one in the first 
angle and another in the third. 

v' 3 2 

X=x' rf- - = — = — 2 as x' = ± oo ; 

Ax'+x' 2 lii 

x' 

Y = y' ^ — = 77 v»= 2 as x' — ± oo. 

+1 " 
a; 



* x = a as y = oo is read "x approaches a as its limit as y approaches 
infinity, or increases without limit." A variable cannot approach infinity 

as its limit. For example, if y = -, and x = 0, y does not approach infinity 

x 
as its limit ; for, when x is infinitely near 0, y is infinitely large ; but it 
doubles its value while x decreases by half its own value. Hence, as x = 0, 
the difference between oo and y must always be many times as great as y, 
however great y may become. 



148 ' TANGENTS, NORMALS, AND ASYMPTOTES. 

Therefore, the line y ' — x + 2, whose intercepts on the axes of 
x and y are respectively — 2 and 2, is an asymptote to each 
branch. 

2. Examine the conic sections for asymptotes. 

Neither the circle nor the ellipse can have an asymptote, since 
neither has an infinite branch. 

The parabola has two infinite branches, one in the first angle 
and another in the fourth. 

v' 2 

Here X = x' = — x' = — oo as x' = go ; 

P 

, px' y' 
and Y=y' r = — = oo as x' = go . 

Hence the parabola has no asymptote. 

The hyperbola has four infinite branches, one in each angle. 
In this curve 

X = x' J— = — = ± as x' = ± oo ; 



Irx 



J x' 



T-,2 12 7 2 

Y = y' ^— = = ± as x' or ?/' = ± co. 

cry' y' 

Hence the asymptote to each branch passes through the ori- 
gin. To determine the direction of these asymptotes, we have 

-f- t = — - - = ± ■ = ± — as x' = ± oo. 

clx' cry' | a 2 a 



4-5 



b 2 x' 
Since ■ — - is positive for anv point in the first or third angle, 

a 2 y' " b 

and negative for any point in the second or fourth, y = -x is 

the asymptote to the branches in the first and the third angle, and 
y — x to those in the second and the fourth. These asymp- 
totes are evidently the produced diagonals of the rectangle on 
the axes. 

3. Prove that y — — x is an asymptote to each of the two 
infinite branches of \f = a 3 — x 3 . 



ASYMPTOTES. 149 

4. Show that y 2 = ax" has do asymptotes. 

139. Asymptotes Determined by Inspection or Expansion. 

From the definition of an asymptote, it follows that it is a line 

which an infinite branch of a curve approaches indefinitely 

near, but never reaches. From this view of asymptotes, we 

can often determine their equations by inspecting the equation 

of the curve, or expanding one of its members. Thus, in the 

cissoid 

x 3 

y 2 = , y = ± oo as x = 2 a. 

2a — x 

Whence, x = 2a is an asymptote to the two infinite branches of 
the curve ; for they approach indefinitely near, but never reach, 
this line. 

In x = log a y, or y = a x ,x = — cK as y = 0. The axis of x is 
therefore an asymptote to the infinite branch in the second 
angle. 

Again, the equation xy — ay—bx=0 may be put in the 
form, 

bx ay 

y = or x = — £-, 

x — a y — o 

from which we know that x = a and y = b each is an asymptote 
to two infinite branches. 

The method of examining a curve for asymptotes by solving 
its equation for y, and then developing the second member in 
descending powers of x, by Maclaurin's formula or some other 
means, will be illustrated by a few examples.* 

* The following is a brief view of another method of examining a curve 
for asymptotes. For a fuller treatment, see Williamson's Differential 
Calculus, page 240. 

Let the equation of a right line be 

y = ixx+v\ (1) 

that of a curve of the nth degree, 

f(*,y)=0; (2) 

that obtained by substituting ux + v for y in (2), 

*(x) = 0. (3) 



150 TANGENTS, NORMALS, AND ASYMPTOTES. 

Examples. 
1. Examine x" — xy 2 + ay 2 = for asymptotes. 

Here y=±x( ) == ± x 1 1 

* \x — a J \ 

. /., . a . 3 or . 

From the first form of the value of y, x = a is evidently an 
asymptote to two branches of the curve that lie to the right of it. 
From the last form we see that two branches of the curve ap- 
proach infinitely near each of the lines y = ± x ± -, as x = ± go. 

The curve therefore has three asymptotes, each of which is 
asymptotic to two infinite branches. 

Equation (3) is evidently of the nth degree, and its n roots are the 
abscissas of the n real or imaginary intersections of (1) and (2). If two 
roots of (3) be equal, two points of intersection of (1) and (2) will coincide, 
and, in general, (1) will be a tangent to (2). From Algebra, we know that, 
as the coefficients of x n and x n ~ l approach zero as a limit, two roots of (3) 
increase without limit. Hence, if fi and v in (1) have such values as ren- 
der these coefficients 0, (1) has the limiting position of a tangent to an 
infinite branch, or is an asymptote. 

For example, let the curve be 

y s = ax 1 -f x s . ( 1 ) 

Substituting /j.x+ v for y, and arranging the terms, we have 

(/t 8 — l)a: 3 + (3^-a)x 2 +3 / uj/ 2 j:+^ = 0. (2) 

Two roots of (2) become oo, when ^u 8 — 1 — 0, and S/x 2 v — a = 0; that is, 
when fi = 1, and v — \a. Hence, y — x +£ a is an asymptote to ( 1 ) . 

From the theory of equations, and this theory of asymptotes, the fol- 
lowing are obvious conclusions : 

(a) Any asymptote or tangent to a curve of the third degree intersects 
the curve in one, and only one, point. 

(b) Any asymptote or tangent to a curve of the nth degree cannot meet 
it in more than n — 2 points, exclusive of the point of contact. 



ASYMPTOTES TO POLAR CURVES. 



151 



2. Examine y 2 = x? 



&* + ! 



for asymptotes. 



l ~i)M 



= ±Xll -2? 



+ " 






Hence y = ± x are the two asymptotes. 
3. Examine y 3 = a a; 2 — x 3 for asymptotes. 



vl?is. y — — x + 



4. Examine y = c -f- 



(a- ft) 1 



for asymptotes. 



5. Examine ?/ 2 



or 3 + ax 2 
x — a 



Ans. y = c, and x = b. 
for asymptotes. 

^ws. a; = a, and y = ± (a? + a) . 



140. Asymptotes to Polar Curves. If, as0 = f , p=oo, and 
the subtangent od = os, it is evident that sn, 
which is parallel to om, is an asymptote to the 
infinite branch pk. Hence, to examine a polar 
curve for asymptotes, we find from its equa- 
tion the values of which make p = ±oo. If 
the corresponding value of the subtangent is 
finite, the line parallel to the infinite radius 
vector, and passing through the extremity of 
the limiting subtangent, is an asymptote. 




Examples. 



Fig. 37. 



1. Examine the hyperbolic spiral pO = a for asymptotes. 

Here p = - ; hence, when = 0, p — oo, and subtangent = — a. 


The curve therefore has an asymptote parallel to the initial line, 
and at the distance a above it. 



152 TANGENTS, NORMALS, AND ASYMPTOTES. 

2. Examine p cos = a cos 2 for asymptotes. 

Here p = ^L22?L_ ; hence, when = -, p = ± go, and subtan- 
cos# 2 

gent = — a. 

The line perpendicular to the initial line at the distance a to 
the left of the pole is therefore an asymptote to two infinite 
branches. 

3. Examine p 2 cos $ = a 2 sin 3 6 for asymptotes. 

Ans. The perpendicular to the initial line at the origin is an 
asymptote. 

4. Show that the initial line is an asymptote to the lituus 
pV# = a. 



Miscellaneous Examples. 

1. At what angle does y 2 = 10 x intersect x 2 -f- y 2 — 144 ? 

Ans. 71°0'58". 



2. Find the subnormal of the curve y 2 = 2 a 2 log a;. 



Ans. — 



3. Find the equation of the tangent to the curve 

^{x + y) = a 2 (x — y) at the origin. 
Ans. y = x. 

4. Find ad what angle the curve y 2 =2ctx cuts the curve 
x 3 — 3 axy -\-y 3 = 0. 

Ans. 0, -^-7T, cot _1 V4. 

5. Find the normal, subnormal, tangent, and subtangent of 

X X 

the catenary y = - (e a + e ") . 

u 

2x 2x o 

a — V 

Ans. Subn. =-(e a — e a ) : norm. = i_- 
4 V y a 

8U bt. = ay ; tan.= y2 - 

s/tf-a 2 -Vy 2 -a 2 



6. Find the subtangeut of the curve y 



EXAMPLES o 153 

X s 



a — x 

Ans. M^^l 
Sa-2x 



7. Examine y 2 (x — 2«) = X s — a 3 for asymptotes. 

Ans. x = 2a ; y = ±(x-{-a). 

8. Examine y(a 2 — or) = 6 2 (2a? -f- e) for asymptotes. 

Ans. y = 0; x-\-a = ; x — a. 

9. Examine for asymptotes the folium of Descartes, 

x? + y s — 3axy = 0. 

Ans. y = — x — a. 

10. Examine (y 2 — l)y = (x 2 — 4)<b for asymptotes. 

Ans. y = x. 

11 t. • c a 2 (a;-o)(a;-3a) . , , 

11. Examine y* = — * — -*-* - for asymptotes. 

x~ — 2 a# 

^1«.5. a? = 2 a ; a? = ; y = ±a. 

12. Examine the hyperbola p = — * — — — — for asymptotes. 

e cos 0—1 

13. Find the length of the perpendicular from the pole upon 
the tangent to the lituus p V# = a. * r 2a 2 p 

V/o 4 + 4a 4 

14. In the hypocycloid xi -f- yi = at, prove that the portion 
of the tangent intercepted between the axes equals a. 

15. Give the different methods of drawing a tangent to any 
plane curve at a given point. 



CHAPTER XI. 




DIRECTION OF CURYATURE, SINGULAR POINTS, AND CURYE 

TRACING. 

141. Direction of Curvature. A curve is concave upward or 
downward at any point, according as in the immediate vicinity 
of that point it lies above or below the tangent 
at that point. 

When a curve, as ab, is concave upward, it 

is evident that -£, the slope of the curve, in- 

iXJb -jo 

creases as x increases ; hence, — ". the deriva- 

° x dv 

Fig. 38. tive of -£, is positive (§ 72). When a curve, 

dx , 

as cd, is concave downward, — ^ decreases as x increases, and 

—4 is negative. 

dx 2 & 

Hence, the curve y = f(x) is concave upward or downward, 
at any point (x, y), according as — ^ is positive or negative. 

In the polar system, a curve is said to be concave or convex 
toward the pole at any point according as in the immediate 
vicinity of that point it lies on the same 
side of the tangent as the pole, or on the 
opposite side. From the figure, it is 
evident that, when the curve is concave 
toward the pole, p or od increases as p 

increases, and -£ is positive (§ 31). 

dp 

When the curve is convex toward the 
pole, p decreases as p increases, and -J- 




Fig. 



is negative. 



dp 



Hence, a polar curve is concave or convex toivard the pole 
according as — is positive or negative. 



EXAMPLES. 155 

When the equation of the curve is given in terms of p and 0, 
we findp in terms of p by use of (8) of § 136. 



Examples. 

1. Find the direction of curvature of y — x 2 -}-2x -\- 5. 

Here — % = 2 ; hence the curve is concave upward. 
dx 2 

2. Find the direction of curvature of y = a + c(x + b) 3 . 

d 2 v 
Here —^-— 6c(x -\-b) ; hence the curve is concave upward or 
dxr 

downward at (a,*, y), according asa;> or < — b. 

3. Find the direction of curvature of y = x? — Sx 2 — 9 # -|- 9. 
Ans. Concave upward or downward according as x> or <1. 

4. Find the direction of curvature of x = \og a y, and y = sinx. 

5. Find the direction of curvature of the lituus p 2 — a 2 . 

Here &= *- =--£• 

M6ie dO 2 P 2 a 2 ' 

p 2 2a 2 P 



ap 

cW 2 



J 2 ,<¥ V4a*+ P *' 
\r T dfi 2 



• dp = 2a 2 (4,a i - P i ) 
' 'dp (4a 4 +p 4 ) f ' 

Hence this spiral is concave or convex toward the pole at (0,p) 
according as p < or > a V2. 

6. Find the direction of curvature of the logarithmic spiral 
p = a 9 . 

Here p = — 



VI + (loga) 2 d P Vl -f (log a) 2 

Hence the curve is concave toward the pole. 



156 



SINGULAR POINTS. 



Singular Points. 

142. Singular Points of a curve are those which have some 
peculiar property. Such points are : first, Points of Inflexion ; 
second, Multiple Points ; third, Conjugate Points ; fourth, Stop 
Points. 

143. Points of Inflexion. A point of inflexion is a point at 
which a curve changes its direction of curvature. Hence a 
Y tangent at a point of inflexion intersects 

the curve. Thus the tangent at p, a point 

of inflexion, cuts the curve at p. 

At a point of inflexion on y=f(x), 

d 2 v 

—? must evidently change its sign, and 

therefore pass through or oo. Hence, 




Fig. 40. 



if 



dx 2 



be found in terms of x. the roots of 



dx 1 

if ^i 

dx 2 



or oo are the critical values of x to be examined. 



changes its sign as x passes through any one of these 



values, this value is the abscissa of a point of inflexion.' 



Examples. 
1 . Examine x J — Sbx 2 + a 2 y = for points of inflexion. 



rr , , » dhi 6 (b — x) A . , , 6 (b — x) 

The root of — | = — * — - — L = is b ; and — * — - — '- 



dx 2 



ddently 



changes its sign as x passes through b ; hence (b, — _j is a 
point of inflexion. 



2 b 3 



x 
2. Prove that the points in which y = csin- cuts the axis of 

a; are all points of inflexion. 

* On one side of a point of inflexion, the slope of a curve is increasing ; 
and, on the other, it is decreasing : hence a point of inflexion is a point of 
maximum or minimum slope, and the method of finding such a point is 



seen to be that of finding a maximum or minimum of 



iii. 

dx 



MULTIPLE POINTS. 157 

The roots of — ^ = — —sm- = are 0, cm-, 2arr, Sair, etc. As 
dx 2 a" a 

x passes through each of these values, — | chauges its sign; hence 
0, a7r, 2a7r, 3a7r, etc., are abscissas of points of inflexion. 

3. Examine the witch of Agnesi, x 2 y — 4 a 2 (2 a — ?/), for 
points of inflexion. ^„ s . (± |aV3, f «)• 

4. Examine ?/ = for points of inflexion. 

a 2 -f- x 2 

Ans. (0, 0), (a V3, faV3), and (.-a.V8, -faV3). 

144. To test curves given by their po/ar equations for points 

of inflexion, we find the roots of — = or oo. If -±- changes 

dp dp 

its sign as p passes through any one of these critical values, 
this value is the radius vector of a point of inflexion.* Thus, 

in the lituus, (28° 38', aV2) is a point of inflexion; for -£ 

dp 
- dp 

changes its sign as p passes through a V 2, the root of -p = 0. 

(See § 141, Ex. 5.) 

Multiple Points. 

145. A Multiple Point is one through which two or more 
branches of a curve 

pass, or at which 
they meet. A mul- 
tiple point is double 
when there are only 
two branches, triple < —/^ / ^^-^ 
when only three, and «• «. 

J ' Fig. 41. 

so on. 

A multiple point at which the branches intersect (Fig. a) is 
called a Multiple Point of Intersection. 

* A point of inflexion on a polar curve evidently corresponds to a 
maximum or minimum of p. 




158 SINGULAR POINTS. 

A multiple point through which two branches pass, and at 
which they are tangent (Figs. b, c) is an Osculating Point. 

A multiple point at which two branches terminate, and are 
tangent (Figs, d, e) is a Cusp. A cusp or osculating point is 
said to be of the first or the second species, according as the 
two branches are on opposite sides (Figs, b, d) or the same side 
(Figs, c, e) of their common tangent. 

A Conjugate Point is one that is entirely isolated from the 
rest of the real locus. Hence, in an algebraic curve, a conju- 
gate point is a multiple point formed by the intersection or 
meeting, in the plane of the axes, of imaginary branches ; that 
is, of branches lying outside of the plane of the axes. Since 
an odd number of roots of an algebraic equation cannot be 
imaginary, an even number of imaginary branches must inter- 
sect or touch in a conjugate point of an algebraic curve. 



146. From the definitions given above, it follows that, at a 
multiple point of intersection, — " must have two or more un- 
equal real values ; that, at a point of osculation or a cusp it 
must have two equal real values ; that, at a conjugate point on 
an algebraic curve, it must have two or more values which .are 
imaginary, unless the tangents to the imaginary branches at the 
conjugate point lie in the plane of the axes. 

Hence at any multiple point -*- has two or more values. 

dx 

147. If f(x, y) = u = be tlie algebraic equation of a curve 
freed from radicals and fractions ; at any multiple point upon the 
curve, 

du 
dy__dx_0 y du _du _ ^ 
dx du 0' dx dy 

dy 

For, at anv multiple point, — , or the ratio of — to — , must 

dx dx dy 

have two or more values (§ 146) . But, from the form of the 



MULTIPLE POINTS. 159 

du du 
equation of the curve, neither — nor — can contain radicals or 

dx dy 

fractions ; hence their ratio can have two or more values for the 
same values of x and y, only when it assumes the indeterminate 

form — 


148. Examination of a Curve for Multiple Points. To examine 
a curve for multiple points, put its equation in the required form 
f(x, y) = u = Q, and find the sets of values of x and y that will 

satisfy the equations — = and — = 0. Of these sets, those 
J ^ dx dy 

which satisfy the equation of the curve give the points to be 
examined. 

Let (#', y') be one of these points ; then 
du 

c it = - — = 9. 
dx'~ dw~~0' 
dy' 

which is evaluated according to the method of § 83. 

du' 

I. If -sL has two or more unequal real values, (x 1 , y') is in 

(XX 

general a multiple point of intersection. 

II. If -^- has two equal real values, (V, y') is in general 

dx' 
either an osculating point or a cusp. 

III. If all the values of -^- are imaginary, (x', y') is a con- 

dx 

jugate point. 

The following considerations enable us to discover more 
exactly the nature of these points : 

In Case I., if the values of y or — ^ are real for x = x'—h 

dx 

and x = x'-\-h, h being very small, (x\ y') is a multiple point of 
intersection ; if real for neither, (V, y') is a conjugate point at 
which the imaginary branches are parallel to the plane of th§ 

axes. 



160 SINGULAR POINTS. 

In Case II., if the values of y or -^ are real for x = x'—h 

dx 

and x = x'+h, (a?', ?/') is a point of osculation ; if real for only 
one of these values of x, (V, 2/') is a cusp ; if real for neither, 
(x\ y') is a conjugate point. 

In some curves, and especially when — = 00, it is better to 
7 dx' 

dx 
inspect the values of a? or — for y — y'— h and y = y'-\- h. 

dy 

To determine the species of a cusp or point of osculation, 

d 2 v 
find — £, and (x',y') will be of the first or the second species, 

according as the two values of —M- have opposite signs or the 

same sign. Or we may compare the ordinates or abscissas of 
adjacent corresponding points on the branches and on their 
common tangent.* 

* The following is a brief view of another method of examining a curve 
for multiple points. For a fuller treatment, see Williamson's Differential 
Calculus or Salmon's Higher Plane Curves. 

Let the equation of a curve of the nth degree be 

/(*,» = 0; (1) 

that of a right line through the origin, 

y = (ix; (2) 

that obtained by substituting fix for y in (1), 

<p(x)=0. (3) 

If (1) contains no constant term, its locus evidently passes through the 
origin. If (1) contains no constant term nor any term of the first degree, 
two roots of (3) are 0. Hence two points of intersection of (1) and (2) are 
at the origin; and the origin is a double point. If, in addition, (1) con- 
tains no term of the second degree, three roots of (3) are 0, and the origin 
is a triple point. 

Hence, when the origin is a multiple point, this fact is evident from the 
equation of the curve. To examine a curve for multiple points not at the 
origin, change the reference of the locus to new parallel axes, using the 
formulas x = m -f x v and y = n -f- y v If m and n can be so determined that 
the resulting equation will contain no constant term nor any term of the 
first degree in x x or y x , (m, n), or the new origin, is a double point. If m 
and n can be so determined that the new equation will contain no constant 



EXAMPLES. 161 



Examples. 



1. Examine x* + a ^V — a V* = f° r multiple points. 

Here u = x 4 + a# 2 ?/ — ay 3 = ; (1 



,'. — = 4-oj 8 -+- 2ax#, and — = aar* — 3 ay 2 ; 
c?# dy 

. dy _ 4 X s -h 2 aagy 

(fa 3 ay 2 — ax 2 



(2) 



Placing the partial derivatives equal to zero, we have 

x(2? + ay) = 0, (3) 

and x 2 -3y 2 = 0. (4) 

Solving (3) and (4), we obtain the following three sets of 
values for x and y, 

# = 0, i«V3, and — £aV3, 

2/ = °> — i«i and -{a. 

Only the first set of values will satisfy (1) ; and (0, 0) is the 
only point to be examined. From (2) we have 

dy~\ _ 4 ar* + 2 axy 
dx_ 0j0 Say 2 — ax 2 

Hence the origin is a triple point at which the inclinations of 
the branches are respectively 0, ^7r, and f ir. 

From (1), 

x= ±\ — \ay ± \y s/±ay + a 2 . (5) 

From (5) , we see that four values of x are real when?/ = — h, 
and two when y = + h ; hence each of the three branches passes 

term nor any of the first or second degree in x x or y ly (m, n) is a triple 
point. 

Erom this method, it is evident that a curve of the third degree can 
have only one multiple point ; one of the fourth degree, only two double 
points or one triple point ; one of the fifth degree, only two double points 
or one triple and one double point, 



= and ± 1. § 83. 

0,0 



162 



SINGULAR POINTS. 



through the origin, which is therefore a triple point of inter- 
section. The general form of the curve at the origin is shown 
in Fig. a on page 157. 

2. Examine y' 2 — a? (a 2 —x 2 ) for mul- 
tiple points. 

Ans. (0, 0), Fig. 42, is a double 




Fig. 42. 

3- Examine x s 



point of intersection 



dx 



= ±a. 



0,0 



3 axy -f if — for multiple points. 



Ans. (0, 0) is a double point of intersection ; -$- 

dx 

4. Examine y 2 (a 2 — x 2 ) = x A for multiple points. 

dy 

dx 



= and oo. 



= ±0, 



and from its equation we see that the curve extends through the 
origin, and is symmetrical with respect to the axis of x\ hence 
the origin is a point of osculation of the first species. 



5. Examine y 2 = a 2 x s for multiple points. 

&~| =±0, 

dx^ o, o 

and from its equation we see that the curve consists of two 
branches symmetrical with respect to the axis of x, and extend- 
ing from the origin to the right ; hence the origin is a cusp 
of the first species, the common tangent being the axis of x 
(Fig. 43). 



6. Examine y 3 = ax 2 — x 3 for multiple points. 

Here (0, 0) is the only critical point. — " 

clx 



= ±oo, and 



y = (ax 2 — ar 5 )? shows that there is a branch on each side of the 
axis of ?/, neither of which extends below the origin, which is 
therefore a cusp of the first species. 

7. Examine y 2 = x (x -f a) 2 for multiple points. 



Here ( — a, 0) is the critical point, 
hence ( — a, 0) is a conjugate point, 



and 



dy 
dx 



= ±0. 

0,0 



CURVE TRACING. 163 

8. Examine a 3 y 2 — 2 abafy — x 5 = for singular points. 

Here (0, 0) is the critical point, and — 

When x = + h, 

bh 2 , \b 2 W + ah 5 
or \ or 
in which both the values of y are real, one greater and the other 
less than 0. 

When x = — h, 



bh 2 , /&%* - oh 5 
or \ cc 



m which both the values of y are real and greater than when 
li is small. 

Hence the origin is a point of osculation, and a point of in- 
flexion on one branch. 

149. A multiple point at which two or more branches termi- 
nate, and have different tangents, is a Shooting Point. A Stop 
Point is a point at which a single branch of a curve terminates. 

From the law of imaginary roots of algebraic equations, 
neither a shooting point nor a stop point can occur on an alge- 
braic curve. 

150. Curve Tracing. The most rudimentary method of trac- 
ing a curve is to find from its equation such a number of its 
points that, when located, these points will clearly indicate the 
form of the curve. 

This method is laborious ; and our present object is to utilize 
the principles heretofore developed to determine directly from 
its equation the general form of a curve, especially at such points 
as present any peculiarity, so that the curve may be traced with- 
out the labor of the first method. 

To trace a curve from its rectilinear equation, the following 
general directions will be found useful. 

Solve its equation for y or a?, and determine any lines or points 
with respect to which the curve is symmetrical. 



164 



CURVE TRACING. 



Find the points at which the curve cuts the axes, and deter- 
mine its limits and infinite branches. 

Determine the positions of its asymptotes, and on which side 
of each the infinite branches lie. 

Find its maxima and minima ordinates, and the angles at 
which it cuts the axes. 

Determine its direction of curvature, points of inflexion, and 
multiple points. 



Examples. 
1. Trace the curve whose equation is y 2 



a 2 x 3 . 



Here y = ± aa$, and the curve is symmetrical with respect to 
the axis of x. 

When x = 0, y — 0, and the curve meets the axes at (0, 0) . 

When x < 0, y is imaginary ; but when x > 0, y is real. Hence 
Y / there is one infinite branch in the first angle, 

and another in the fourth. 




dy = 
dx 


3 a 2 x 2 , 3 axi . , 

= = ± = oo when x = go 

2y 2 


hence the curve has no asymptote. 


dy 3a 2 x* , ' /A AN 

-£■ = = - at (0, 0) ; 

dx 2y 


but Sf 

dx 


_ , 3 ax^ 

0,0 " 


= ±0. 

0,0 



Fig. 43. 



Hence the two symmetrical branches terminating at the origin 
are tangent to the axis of x at that point, and the origin is a 
cusp of the first species. 
3a 

dx 2 



d u 
Since —4 



, the upper branch is concave upward, and 



4Vx 
the lower one concave downward. The form of the curve is 

shown in Fig. 43. 

2. Trace the curve y z —2 ax 2 — x 3 . 
Here y = (2 ax 2 — X s ) K 

For y = 0, x = and 2 a ; hence the curve cuts the axis of x 
at the origin, and 2a at the right of it. 



EXAMPLES. 165 

For each real value of x, y has one, and only one, real value, 
which is + or — according as x < or > 2 a. Hence there is 
one infinite branch in the second angle and another in the 
fourth. 

To find the equation of the asymptote, we have 
y = (2 ax 2 -x s )h = -x(l- — Y 

= — xll _...) (1) 

V 3a 9a 2 J w 

when x is numerically greater than 2 a. 

Hence the equation of the asymptote to each infinite branch is 

y =--* + !<*. (2) 

From (1) and (2), it is evident that the infinite branches lie 
between the asymptote and the axis of x. 

dy 4: ax — 3 a 2 n , A 

-^ = = when x = 4a ; 

da 3/ 3 

/. (2ax 2 -x")% a , orfa^/4, 
is a maximum ordinate. 



ay 

da 



= 00 
2a, 



hence the curve is perpendicular to the axis of x at (2 a, 0) . 

dy ^ax — Sx 2 , n 

— = = - , when x = y = 0. 

dx 3z/ 2 * 

By evaluating we find that 



dy 
dx 



= ± oo ; 

0,0 



hence, as the curve does not extend below the axis of x to the 
left of x = 2 a, and y has one, and but one, real value for each 
value of a, the origin is a cusp of the first species, the two 
branches being tangent to the axis of y. 



166 



CURVE TRACING. 



c^/ = -8a 2 
dx 2 9x-3(2a-x)i' 

which is + or — , according as x > or < 2 a ; hence (2 a, 0) is 

a point of inflexion, to the right 
of which the curve is concave 
upward, and to the left down- 
ward. The form of the curve 
is given in Fig. 44. 




Fig. 44. 



3. Trace the curve 

y 2 (x 2 — a 2 ) — x*. 

Since its equation involves 
only even powers of x and y, 
the curve is symmetrical with 
respect to each axis. Hence, if 
we determine the part of the locus that is in the first angle, 
the symmetrj- of the curve will give us the other three parts. 

Solving for y, we have 

x 2 
y (x 2 -a 2 )$ 

When x = 0, y = ; but, for other values of x between —a 
and + a, y is imaginary ; hence (0, 0) is a conjugate point, and 
the locus in the first angle lies to the right of x= a. 

As x = a from a value greater than a, y == oo ; hence there is 
one infinite branch in the first angle to which x — a is the as} r mp- 
tote. When x = oo, y = oo ; and there is a second infinite branch 
in the first angle. 

To find the equation of the other asj-mptote, we have 

x 2 



y = ±. 



x 1 



= ± X 



= ±x 



>-f 



a? 

1 + — + 

2x 2 



when x is numerically greater than a ; hence y = x is the equa 
tion of the other asymptote. 



EXAMPLES. 



167 



Evidently the curve lies above this as3'mptote. Hence the 
branch in the first angle lies above y = x, and to the right of 
x = a. 

dy __ 2x 3 -xif _ n 



y(x 2 -a 2 ) 

when ?/ 2 = 2ar, or a; = ± aV2 ; 
hence 2 a is a minimum ordi- 
nate. 



In the first angle, 



cl 2 y == a 2 (x 2 + 2a 2 ) 



dte 2 



(ar^-a 2 )! 




Fig. 45. 

The form of the 



As this is + when x > a, this 
branch in the first angle is concave upward, 
curve is given in Fig. 45. 

4. Trace the curve—- -f- ■*— = 1. 

a s b 3 

The curve cuts the axis of x at (a, 0) and the axis of y at 
(0, b) . There is one infinite branch in the second angle and 

another in the fourth, y = a? is the equation of the as} r mp- 

a 

tote, which lies below the infinite branches. The curve is con- 
cave upward, except between x = and 
x = a, where it is concave downward. 
(0, b) and (a, 0) are points of inflexion. 

x 

5. Trace the curve y = -• 

1 + ar 

The curve has one infinite branch in the 
first angle and another in the third, to each 
of which the axis of x is an asymptote. 
£ is a maximum, and -J a minimum, 
ordinate. (0, 0), (- VB, - J-V3), and 
(V3, iV3) are points of inflexion. The 
inclination of the curve at the origin is \tt. 

a 4-x 




Fig. 46. 



6. Trace the curve y 2 = 
see Fig. 46. 



For the form of the curve, 



168 CURVE TRACING. 

151. Tracing Polar Curves. When possible, write the equa- 
tion in the form p =f(0) . Solve f(0) = to find the angles at 
which the curve cuts the polar axis at the pole. Assign to 
such positive and negative values as make p easily ' found. 

Solve -£ = to find the values of for which p is a maximum 

dO > H 

or minimum, and for which the curve is perpendicular to the 
radius vector. Examine the curve for asymptotes, direction of 
curvature, and points of inflexion. The facts thus obtained will 
indicate the form of the curve. 



Examples. 

1 . Trace the curve p = a sin 3 0. 

Since p = a sin 3 0, p reaches its maximum value a when 
sin30= 1 ; that is, when = \ir, |-7r, f 77- , etc. ; and p reaches 
its minimum — a when sin 30 = — 1 ; that is, when = |-7r, %ir, 
-LL-tt, etc. 

Since -£=3<xcos30, p increases from to a, while in- 
creases from to \tt ; p decreases from a to 
— a, while 6 increases from -J-7T to J-w; p 
increases from — a to -fa, while 6 increases 
from \ir to f 7T ; and p decreases from a to 
0, while 6 increases from -f tt to ir. Further 
revolution of the radius vector in either 
Fj g- 47 « direction would evidently retrace the three 

loops already found. The curve is that represented in Fig. 47. 

2. Trace the curve p = a sin 20. 

The curve consists of four loops. From this and the previous 
example, we infer that the locus of p = asin?i0 consists of n 
loops when n is odd, and 2 n loops when n is even. 

3. Trace p = a sin — 4. Trace the lituus p = 

5. Trace p = a cos -}- 6, in which a > 6. 




MISCELLANEOUS EXAMPLES. 169 



Miscellaneous Examples. 

1. Examine x 4 — axy 2 — ay 3 = for multiple points. 

The origin, a triple point, is a cusp of the first species, through 
which a branch of the curve passes. 

2. Examine ax 3 -f- by 3 — c = for points of inflexion. 

3. Prove that (0, 0) is a multiple point of intersection on the 
curve x* — a 2 xy + b 2 y 2 = 0. 

4. Examine x A — a 2 x 2 -f- a 3 y = for points of inflexion. 

An*. fJL.*«\ a .nd f-JL.Sa 



VV6 36/ V V6 36 

5. Examine ay 2 — x 3 — bx 2 = for multiple points. 

6. Examine ay 2 — x 3 -f- bx 2 = for multiple points. 

Ans. (0, 0) is a conjugate point. 

7. Trace the curve y 2 = — — — 

J x+b 

8. Trace the folium of Descartes y 3 — 3 axy + sc 3 = 0. 

9. Trace (?/-^) 2 = ^. 

10. Trace y 2 (x — a) = cc 3 . 

1 1 . Trace p cos = a cos 2 0. 




CHAPTER XII. 

CURVATURE, EYOLUTES, ENVELOPES, AND ORDER OF 

CONTACT. 

152. If <f> represent the inclination of any curve ab referred 
to rectangular axes, then <£ will measure the direction of the 
curve with respect to the axis of x. At the 
point p, <£ = angle xor ; and at p', </> = xmp' ; 
hence angle prm = A<£, if arc pp'= As, s repre- 
senting the length of the curve. 

153. The Curvature of a curve at any point 
To 7m~~ a is the rate of change of its direction relative 

Fig. 48. to that of its length. 

Hence, if k represent the curvature of any curve, 

K = ^,or i^r^l §31. 

ds As = °|_AsJ 

154. Curvature of a Circle. If ab (Fig. 48) be the arc of a 
circle whose radius is r, the angle prm equals the angle sub- 
tended by the arc pp' at its centre ; and from § 40 we have 

, arc pp' A(f> 1 

angle prm = , or — -*- = - ; 

r As r 

.-.£*, or k, =-. §17, Cor. 2. 

ds r 

Hence, the curvature of any circle is equal to the reciprocal of 
its radius; and the curvatures of any tiuo circles are inversely 
proportional to their radii. 

Cor. Ifr = l, k = 1 ; that is, the unit of curvature is the cur- 
vature of a circle whose radius is unity. 



RADIUS OF CURVATURE. 



171 



155. To find k in terms of the differentials of x and, y. 

tan cf> = — ? ; 

dx 

2 , 7 , cZ 2 ?/ 
.*. sec-<£ a<£ = — - ; 
c?# 



dcf> = 



civ 

dx 



1 + 



c/0" 
ds 



since sec 2 <£ = 1 -f- 
dx 2 



sdxj 



1+f-r- 



aY 2 

dXj 



We take the positive value of the radical so that the curva- 

d 2 v 
ture of a curve will be positive or negative according as — - is 

dx? 

positive or negative ; that is, according as the curve is concave 
upward or downward. The sign of curvature, however, is often 
neglected. 



156. Radius of Curvature. As the radius of a circle varies 
from to oo, its curvature varies from go to ; hence there is 
alwa}~s a circle whose curvature is equal to that of any curve at 
any point. A circle tangent to a curve, and having the same 
curvature as the curve at the point of contact, is called the 
Circle of Curvature of the curve at that point. Its radius and 
centre are the radius of curvature and the centre of curvature of 
the curve at that point. Hence, if R represent the radius of 
curvature of a curve, from §§151 and 155, we have 



K 



1 + 



aYv 

dx) 



dry 
dx 2 



R will be positive or negative, according as the curve is con- 
cave upward or down ward ; but its sign is often neglected. 



172 CTJEVATURE. 

157. The radius of curvature in terms of polar coordinates 
can be found by transforming the value of R in § 156 to polar 
coordinates. We thus obtain 



P dO 2 ' dO 2 dO 2 l dO 2 

in which N is the normal. See § 131, Ex. 3, and § 136, (7). 

158. The circle of curvature, in general, cuts the curve at the 
point of contact. 

For, on one side of the point of contact, the curve changes 
its direction more rapidly than the circle of curvature, and hence 
lies within the circle ; while on the other side it changes its 
direction more slowly than the circle, and hence lies without the 
circle. 

159. At a point of maximum or minimum curvature,* the cir- 
cle of curvature does not cut the curve; and conversely. 

For, on either side of a point of maximum curvature, the curve 
changes its direction more slowly than at this point ; hence, on 
each side of this point, the curve lies without the circle of curva- 
ture at this point. On either side of a point of minimum 
curvature, the curve changes its direction more rapidly than at 
this point ; hence, on each side of this point, the curve lies 
within the circle of curvature. 

Since the conic sections are symmetrical with respect to nor- 
mals at their vertices, it follows that their vertices must be points 
of maximum or minimum curvature. 

Examples. 
1 . Find the curvature of the parabola y 2 = 2px. 

Here -^=-2-, and — %. = — ^ ; 

dx y dx- y 3 

* By a maximum or minimum curvature, we mean a numerical maxi- 
mum or minimum, the sign of curvature not being considered. 





EXAMPLES. 

dx 2 p 2 f y 2 




[ i+ &"] 


i tf\y z +p\ 



173 






(y 2 +P 2 Y 



p 2 . 

The upper or lower sign is to be taken according as — ~ is 
positive or negative. 

At the vertex (0, 0), k = -, which is evidently the maximum 
curvature of the parabola. ^ 

2. Neglecting its sign, find the curvature of the ellipse 

a 2 y 2 + b 2 x 2 = a 2 b 2 . 

TT dy b 2 x , d 2 y 5 4 

Here -^- — — , and— ^ = — ; 

dx a~y dxr cry 5 

" * a 2 f \a,y + Vx 2 ) (a 4 y 2 + Vx 2 ) I 
At the vertex (a, 0) , k = — ; at the vertex (0, — b) , k = — ; 

ft" 

hence, the maximum curvature of the ellipse is — - , and the 

b b ' 

minimum-—- 
a 2 

3. Find the radius of curvature of the cycloid 

x = r vers - — ^J2ry — y 2 . 



Here ^ = V2ry-^ ^ <%L = _2_, 

dx y dx- y z 

which equals numerically twice the normal. 

1 

k = ; 

— 2^/2 ry 

and , the curvature at the highest point, is evidently the 

minimum curvature of the cycloid. 



174 



EVOLUTES. 



4. Find the curvature of y = x i — 4# 3 — 18 x 2 at the origin. 
Find the abscissas of the points at which the curvature is 0. 

Ans. k = — 36 ; x = 3 and — 1 . 

5. Find the curvature of the logarithmic curve y = a x . 



Ans. 



my 



6. Find the numerical value of the radius of curvature of the 
cubical parabola y s = a 2 x, , Q 4 4 x 3 

Ans. B= Vy+ a > T . 
Qa'y 

7. Find the radius of curvature of the spiral of Archimedes 
p = ad. 

Here ^=a, and ^ = 0; 



.R 



(*+£ 



2 Y 



I (i±tf)i_ n (±±f)i 



,d P ^_ d 2 - 

32 P 



p 2 + 2 ^tL P 2 + 2a 2 



2+0 2 
dO 2 ' dO 2 

8. Find the radius of curvature of the logarithmic spiral 
p==ae ' Ans. R = P Vl+(loga) 2 . 

E volutes. 

160. The Evolute of a given curve is the locus of the centre 
of curvature of the curve. The given curve is called the 
Involute of its evolute. 

161. To find the equation of the evolute of any given curve. 
Let c, (a, f3) , be the centre of curvature of the curve ab at any 

point p, (x,y). Then. 

since 



bp = pc sinBCP = R sinKPH = R— , 

ds 




and 



and 



bc = R cos bcp = R 



dx t 
ds ' 

a = oe — bp = x — R-, 
ds 



= ep + bc = y + R 



dx 
ds 



(1) 

(2) 



EXAMPLES. 



Substituting in (1) and (2) the values of R and ds, we have 



x — 



1 + 



d£\dy 
dx 2 } dx 



1 + 



and 



P = y 



dx 2 
dx 2 



dry 
dx 2 



(3) 



W 



By differentiating the equation of any given curve, and sub- 
stituting the results in (3) and (4), a and (3 may be expressed 
in terms of x and y. If, between the equations thus obtained 
and that of cue given curve, x and y be eliminated, the resulting 
equation between a and /5 will be the equation of the evolute. 



Examples. 
1. Find the equation of the evolute of the parabola. 



Here *=?, 



and -1 = 

dx 2 



P- 



dx y J dx 2 y 3 

Substituting these values in (3) and (4) of § 161, and reducing, 
we have 

a = 3x-\-p, or x = a ~-P , 
o 

and fi = — y or y = — &pi. 

Substituting these values of x and y 
in the equation of the parabola, we 
have 

which is the equation of the evolute of 
tf = 2px. Fig - 50 

bac (Fig. 50) is the evolute of the parabola mon. 




176 EVOLUTES. 

2. Find the equation of the e volute of the ellipse. 
Here 



ty _ _ 2?' and ^ = - — ; 
da? a 2 ?/ dx 2 a 2 y^ 



(gt-b 2 )*? 



, or x = 



a 4 a 
^ 2 -5 2 



and 



(a 2 -5 2 ) 2/ 3 



_ / & 4 ff V 
^a 2 - &y 



Substituting these values of a? and y in «y + b 2 x 2 = a 2 6 2 , we 
obtain as the equation of the evolute of the ellipse, 

(aa)* + (ty3)i = (a 2 -&*)l. 



3. Find the equation of the evolute of the cycloid 



x — r vers 



ry — y% 



(1) 



Here 



dy _ -\/2ry-y 2 , d 2 y _ 



dx 






.-. 2/ = - /?, and a; = a - 2 V- 2 r£ - £ 2 



Substituting these values of x and 2/ in (1), we obtain as the 
equation of the evolute of the cycloid, 

.-i/^A^ y_ 2 r/?-/3 2 . 



a = r vers 



(2) 



The locus of (2) is another cjxloid equal to the given cycloid, 

the highest point being 
at the origin. For, if 
Xl the cycloid oo 2 m be 
referred to the axes 
o^ and 0^, and p be 
any point, 



y = - bp, 

and x = OjB 

= ke + dp. (3) 
em = arc ep ; 




PROPERTIES OF THE E VOLUTE. 177 



.*. ke = arc ap = r vers -1 — = r vers -1 1 — - )• (4) 



dp = Vad-de = V— ?/(2?' + y) = V— 2ry — y 2 . (5) 
From (3), (4), and (5), we obtain 

x = r vers -1 ( —1 ] -f-V — 2ry — y' 2 . (6) 



Comparing (2) with (6), we see that os, the evolute of oo 15 
must be equal to o^i ; that is, the evolute of a cycloid is an equal 
cycloid. 

Properties of the Evolute. 

162. Any normal to the involute is a tangent to the evolute. 
The equation of the normal to y =f(x) at (x' f y') is 

y-y'=~(*-*')- (i) 

Let (a, /3) be the centre of curvature of y = f(x) at (V , y') ; 
then (1) passes through (a, /?), and we have 

2/ '-^=-S ( "''" a); (2) 

.•. a ,'_a+g!( 2 ,'- / 8)=0. (3) 

If (x', y') move along the involute, (a, j3) will move along 
the evolute, and a, /?, and y' will be functions of x'. Differen- 
tiating (3) on this hypothesis, we have 

dx < _ da + dy"-dfdfi + (y. _ ff) it = p. 

ax ax 

Dividing by dx 1 , and rearranging terms, we have 

T <fc' 2T ^ ^cte' 2 da;' da;' 2 V ; 



178 EVOLUTES. 



But, as (a, /?) is on the evolute, we have by § 161, (4), 



eto' 2 



dx' 2 

■■^-"ffg+'+S--* ( 5 ) 

From (4) and (5) , we obtain 

__ da _ dy'dfi _ dx* _ d/3 

dx' dx 12 ' dy'~ da 

Hence (1) is tangent to the evolute at (a, /?). 

163. Any continuous arc of the evolute is equal to the differ 
ence between the radii of curvature of the involute that are tangent 
to this arc at its extremities. 

If (x — a) 2 + (y — j3) 2 = R 2 be the circle of curvature of any 
curve at (x', y 1 ) , we have 

(x'-ay+(y>-py = n 2 . (i) 

Suppose (x f , y') to move along the curve ; then y 1 , a, /?, and 
R will be functions of x\ 

Differentiating (1) on this hypothesis, we obtain 
O' - a) dx' + (y' - /?) dy' - (x'- a) da 

-(y'-f3)d(3 = RdR. (2) 

From equation (2) of § 162 we have 



,_ £=- 1^'- .); 


(3) 


da 


(4) 


From (3), 




(V- a)da;'+ (?/'- /8)cfo/'= 0. 


(5) 


From (2) and (5), 




(V- a)da + (2/'- 0)d0 = - RdR. 


(6) 



PROPERTIES OF THE EVOLUTE. 



179 



From (1) and (4), 



(?) 



(8) 



da 2 
From (4) and (6), 

da 

Squaring (8) and dividing by (7), we obtain 
da 2 + cZ/5 2 = dPi 2 . 

But, s being the length of the evolute, we have 

da 2 + dp 2 = ds 2 ; 
.*. ds = ± dU ; 
that is, R increases or decreases as fast as s increases. 



Hence, if the length of the evolute of the parabola (Fig. 52) 
be estimated from the point a, we have 



arc ap = kp 



oa. 



Again, if the length of the evolute of the cycloid (Fig. 51) 
be estimated from o, 

arc os = so! = 4r ; 

hence the length of one branch of the 
cj^cloid is 8 r. 

164. These two properties of the 
evolute enable us to regard any invo- 
lute as traced by a point in a string 
unwound from its evolute. Thus, if 
on a pattern of cpao one end of a 
string be fastened at c, and the string 
be then stretched along the right of 
cpa, the point of the string which 
reaches o, when carried around to the right, will trace the arc 
ok??i as the string unwinds from the evolute. 




Fig. 62. 




Fig. 53. 



180 ENVELOPES. 

Since any point of the string beyond a will trace an involute 
of ca, it follows that, while a curve has but one evolute, it can 
have an infinite number of involutes. 



Envelopes. 

165. If, in the equation f(x, y, a) = 0, a series of different 
values be assigned to a, the equation will represent a series of 

curves differing in form, or in 
position, or in both these re- 
spects, but all belonging to 
the same class or family of 
curves. 

For example, if different 
values be assigned to a in 
(x — a) 2 + y 2 = 25, its loci will 
be a series of equal circles with their centres on the axis of x 
(Fig. 53). 

The quantit}' a, which is constant for any one curve, but 
changes in passing from one curve to another, is called a Vari- 
able Parameter. Any two curves of a series that correspond 
to nearly equal values of the parameter usually intersect, and 
are called consecutive curves. 

166. An Envelope is the locus of the limiting positions of the 
points of intersection of the consecutive curves of a series, as 
these curves approach indefinitely near each other. 

167. To find the equation of the envelope of a series of curves. 
Let f(x, y, a) = u = (1) 

and f(x, y, a -f- Aa) = (2) 

be the equations of any two consecutive curves of a series ; then, 
at the points of intersection of (1) and (2), we evidently have 

f(x, y, a + Aa) —f(x, y, a) __ Q> ^ 

Aa 



ENVELOPES. 181 

Passing to the limit as Aa = 0, we have, at the limiting posi- 
tions of these intersections, 

■f/(*,y,a) = 0, or ff = 0. (4) 

(La Cla 

Since the coordinates of the points on the envelope satisfy 
both (4) and (1), its equation is found by eliminating a between 
these equations. 

168. The envelope is tangent to each curve of the series. 
Let cj> (a?, y) represent the value of a obtained from (4) of 
§ 167; then, if a =<£(<», y), 

f(x,y,a)-u=0 (1) 

is the equation of the envelope. 

Differentiating (1), a being variable, we obtain 

du 7 . die , . du , A 

— dx-\ ay -\ da = 0. 

dx dy da 

But, at any point on the envelope, 

^ = 0; §167, (4) 

da 

.'.^clx + — % = 0, (2) 

ax ay 

which gives the slope of the envelope at an} T point. 

Differentiating /(a?, y, a) ==. u = 0, considering a constant, we 
have , 7 

<^dx + ^dy = 0, (3) 

dx dy 

which gives the slope of any individual curve at any point. 
From (2) and (3) we see that the envelope and any curve of 
the series have the same slope at their common point. 



Examples. 

1. Find the envelope of (x — a) 2 -f- y 1 — r 2 , in which a is a 
variable parameter. 



182 ENVELOPES. 

Here f(x, y, a)[= u\ = (x — a) 2 -\-y 2 — r 2 = ; (1) 

.: — = -2(x-a) = 0. (2) 

da 

From (1) and (2), y == ± r ; that is, the envelope is two lines 
parallel to the axis of x, as would be inferred from Fig. 53. 

2. Find the envelope of y = ax-\ — , a being the variable 
parameter. 

Here /(a?, ?/, a)[= u]=y — ax =0; (1) 

a 

.\—=-x + ^ = 0. (2) 

da a 

Eliminating a between (1) and (2) , we obtain y 2 = A,mx ; that 
is, the envelope is a parabola whose latus rectum is 4m. 

3. Find the envelope of the hypotenuse of a right-angled 
triangle of constant area. 

Let a and j3 be the sides of the right triangle, and assume 
them as coordinate axes ; then 



a (3 
is the equation of the Irypotenuse. 
Let c = the constant area ; then 

a/? = 2c, or j3 = — • 
a 

Hence f(x, y, a ) = ^ + ^- 1 = 0, (1) 

a A C 

and ^ = __^_ + J/- =: o. (2) 

da a 2 2c W 

Eliminating a between (1) and (2), we obtain xy==^c; that 
is, the envelope is an hyperbola to which the sides of the tri- 
angle are asymptotes. 



CONTACT OF DIFFERENT ORDERS. 183 

4. Find the envelope of a system of concentric ellipses, the 
area and the directions of the axes being constant. 

Let the equation of the ellipses be 

a y + /5V = a 2 /3 2 , (1) 

and c represent the constant area ; 

then c = 7ra/?. (2) 

Eliminating j3 between (1) and (2), we have 

f(cc, y, a) = o.y + ^-^ = 0; (3) 

TT a 77 

lu =2ay 2 -^4^ = 0. (4) 



d< 



a 



From (3) and (4) , we obtain 

xy = ± — -, 

9 2tt 

which are the equations of conjugate equilateral hyperbolas 
referred to their asymptotes. 

169. Contact of Different Orders. Let y—f(x) and y=4>(x) 
be an} T two curves referred to the same axes. If f(a) = <£(«), 
the curves have the point [«,/(«)] in common. If f(a) = <j>(a) 
and /'(a)= <£'(a), the curves are tangent at [a, /(a)], and are 
said to have a contact of the first order. If f(a) = $(a) , f'(a) 
= <£'(a), and /"(<*) = <f>" (a), the two curves have the same 
curvature at their common point, and their contact is of the sec- 
ond order. If in addition, /'"(«)= </>"'(a), their contact is of 
the third order; and so on. Thus, contact of the nth order im- 
poses n + 1 conditions, 

170. Two curves intersect or do not intersect at their point of 
contact, according as their order of contact is even or odd. 

Let y =f(x) and y=cf>(x) be sany two curves having the point 
[a, /(a)] in common. Let h be a very small increment of x 
By Taylor's formula, we have 



184 OEDEB OF CONTACT. 

A«+*)=A«) +/(«)*+/'(«) i^+/"'(«)'ir + - W 

«a + *) = +(a) + *'(»)* +^(a)i^ + *"'(o)i^ + ...(2) 

If. Li. 

Subtracting (2) from (1), we obtain 
f {a + h) -^a+h)=h[f\a)-<t,\a)^ + ¥-U«(a)-4,»(ay] 

which gives the difference between the corresponding ordinates 
of the curves on each side of their common ordinate. If the 
contact is of an odd order, the first term of the second member 
of (3) which does not vanish contains an even power of h ; 
hence the sign of the second member is the same whether h be 
positive or negative. Therefore, y=zf(x) lies either above or 
below y = </>(#) on ^°^ n s ^ es of their common point; and the 
curves do not intersect. But, if their contact is of an even 
order, the first term of the second member of (3) that does not 
vanish contains an odd power of h. Hence, in this case, the 
second member changes sign with h ; and y=f(x) lies above 
y = (/>(#) on one side of their common point, and below it on the 
other ; and the curves intersect. 

Cor. At a point of maximum or minimum curvature, the 
circle of curvature has contact of the third order with the curve ; 
for it does not cut the curve at such a point. 

The following are obvious conclusions from equation (3) : 

(a) Two curves on each side of their common point are 
nearer, the higher their order of contact. 

(6) If two curves have a contact of the nth order, no curve 
having with either of them a contact of a lower order can lie 
between them near their common point. 

171. Osculating Curves. The curve of a given species, that has 
the highest order of contact possible with a given curve at any 
point, is called the osculating .curve of that species; 



OSCULATING CURVES. 185 

Let y =/(#) be the most general form of the equation of a 
curve of a given species, and suppose that it contains n-f-1 
arbitrary constants. Upon the n +1 constants, n -f-l 5 and only 
n+1, independent conditions can be imposed. But, in order 
that y=f(x) may have contact of the ?ith order with a given 
curve at a given point, n +1 conditions must be fulfilled by its 
constants ; that is, these constants must have such values that 
y=f(x) shall pass through the point, and the first n derivatives 
of its ordinate be equal to those of the given curve at this point 
(§ 169). 

Hence, as y = ax + b has two constants, the osculating straight 
line has contact of the first order, and is a tangent. 

As (x — a) 2 + (y — b) 2 = r 2 has three constants, the osculating 
circle has, in general, contact of the second order, and is the circle 
of curvature. 

The osculating parabola has contact of the third order. T7ie 
osculating ellipse or hyperbola has contact of the fourth order. 



Miscellaneous Examples. 
1. Find the curvature of the hyperbola. 

Ans. K =± 



(6V + aY)i 



2. Find the radius of curvature of the equilateral hyperbola 

Xy=W - Ans. B = ±.^ + yy. 

a 2 

3. Find the radius of curvature of y 3 = 6 x 2 -f x 3 . 

Ans. B = C^ + ^ + s 2 ) 2 ]* numerically. 
Sx-y 

4. Find the radius of curvature of the lemniscate of Ber- 
nouilli,p 2 =« 2 cos 20. at?® 2 

5. Find the curvature of the cissoid y 2 = - 



Ans. k = ± 



f — x 

3(2a-x) 2 



aa55(8a — £x)\ 



186 INVOLUTES AND ENVELOPES. 

6. Find the equation of the evolute of the hyperbola. 

Ans. (aa) !-(&/?) f = (a 2 + & 2 ) 1 - 

7. Find the length of ^he evolute of the parabola in terms of 
the abscissas of its extremities. 

B== (1±P1 == ^±P)!. § 159, Ex.1. 

Arc ap (Fig. 52) = kp - oa= ( 2a; -H>)* _ p 

-Vp 

=^(^) ! - §161 - Ex - 1 - 

8. Find the envelope of y 2 =a(x — a), in which a is a 
variable parameter. 

Ans. y = ± \x. 

9. One angle of a triangle is constant and fixed in position ; 
find the envelope of the opposite side when the area is constant. 

Ans. xy — — : — , c being the constant area and w the con- 
2 sin w 

stant angle. 

10. Find the envelope of the circles whose diameters are the 
double ordinates of the parabola y 2 = 2%)x. 

Ans. y 2 == p (p -f- 2 x) . 

11. Find the envelope of a line of constant length a, whose 
extremities move along two fixed rectangular axes. 

Ans. xi -{-y% = a%. 

12. Find the envelope of the normals to the parabola y 2 =2px. 

Ans. y 2 = (x —p) 3 , which is the evolute of the parabola, 

27 p 

as it clearlv should be. 



1 3 . Find the radius of curvature of the catenary y = - ( e a + 



f 

Ans. B = — 

a 



MISCELLANEOUS EXAMPLES. 187 

' 14. Find the radius of curvature of the cardioid p — a{l— cosO). 

Ans. B = {*«&. 
3 

fie ?/ 2 

15. Find the envelope of the series of ellipses ■ — | ^ =1 

a 2 (k—a) 2 
a being a variable parameter. 

Ans. #1 -f y^ = &*• 

16. Prove that the whole length of the evolute of the ellipse 

a 3 - b s 



is 4 



ab 



17. Find the radius of curvature of the tractrix, having given 

^ = -^ §195, Ex.5. 4ns. 2? = - (a 8 - y 2 ) 4. 

cfo Va 2 -2/ 2 y 

18. Show that the evolute of a circle is its centre. 

af - — -\ 

19. Prove that the catenary /?=-j e a — e « j is the evolute 

of the tractrix a; = a log — — —^-{a 2 -y 2 )h. §201, Ex.1. 

20. Find the equation of the envelope of y=ax± (<xV-f fr 2 )*, 
in which a is a variable parameter. 

Here the equation of the tangent is given, and that of the 
curve is required. * x 2 y 2 _ . 

US ' ~tf ¥~ 

21. Prove that — + '— = 1 is the envelope of the circles 

a 2 + b- b 2 

described on the double ordinates of the ellipse a 2 y 2 -\-b 2 x 2 =a 2 b 2 
as diameters. 



CHAPTER XIII. 

INTEGRATION OF RATIONAL FRACTIONS. 

172. Decomposition of Rational Fractions. Any rational frac- 
tion whose numerator is not of a lower degree than its denomi- 
nator can be separated by division into two parts, the one 
rational and entire, and the other a rational fraction whose 
numerator is of a lower degree than its denominator. For 

example, 

x* , 5 x 2 — 4 

x — 2 



x" -j-2x 2 — x— 2 x* -\-2x 2 — x — 2 

Hence airy rational differential can be considered as com- 
posed of an entire part and a fraction whose numerator is of a 
lower degree than its denominator. The entire part can be 
integrated by previous methods ; and it is our present object to 
show that the fractional part, if not directly integrable, can be 
resolved into partial fractions which are integrable. These par- 
tial fractions differ in form, according as the simple factors of 
the denominator of the given fraction are : 

I. Real and unequal. 

II. Real and some of them equal. 

III. Imaginary and unequal. 

IV. Imaginary and some of them equal. 

To show, in the simplest manner, how the decomposition and 
integration is to be effected, we shall apply the process to par- 
ticular examples in each of the four cases. 

173. Case I. When the simple factors of the denominator are 
real and unequal, to every factor, as x — a, there corresponds a 

partial fraction of the form 

x a 



CASE I. 189 



Let it be required to find ( \ - ~\ ' — ■ 



+ 

The roots of x 3 -\-x 2 —2x = are 0, 1, and — 2; hence, the 
factors of X s + $? — 2 x are x, x — 1 , and x-\- 2. 

A 2x + 3 A . B C ,-* 

Assume ! = (1) 

x(x—l)(x-j-2) x x-\ x+2 w 

Clearing (1) of fractions, we have 

2x + 3 = A{x - 1) (a; + 2) + 5 (x + 2)# 

+ C y (^-1)^ (2) 

= (^+5+0)a; 2 +(^+2£-C>-2A (3) 

Equating the coefficients of like powers of x in (3) , we have 
A+B+C=0, A + 2B-C=2, and -2A=B. (4) 

Solving equations (4), we find 

A = -l £ = f, andC = -l. (5) 

Substituting these values in (1), we obtain 

2ar+3 _. 3 5 1 (Q , 

a? + x 2 -2x 2x 3(a?-l) 60 + 2) ' K) 

r (2x + 3)dx = 3 Cdx 5 T dx I T eta 
"J x 3 + x 2 — 2x~ 2J x 3J x — 1 6J x-j-2 

= - 1 logx + 1 log (a; - 1) - J log (a + 2) + logc 

t c(cc — 1)§ 

= log— ^ ^ — 

5 xi(a; + 2)i 

Equation (6) is true ; for the values of A, B, and C given in 
(5) satisfy (4), and hence make (3),* and therefore (1), an 
identical equation. 

* The principle used here is : IfA = A', B = B', C = C, etc., A + Bx 
+ Cx 2 + ••• = A'+ B'x + C'x 2 -f ••• is an identical equation. 

That the fraction has been correctly decomposed is proved by reasoning 
backward through the process. Equation (1) is not assumed as a basis of 
proof, but as a basis of operation. 



190 INTEGRATION OF RATIONAL FRACTIONS. 

The values of A, B, and C may be obtained from (2), as 
follows : 

Making x — 0, we have 3 = — 2 A; \ A = — f . 



_ 3 

Making a; =1, we have 5 = 3B; .\B = % 



3' 



Making x = — 2, we have — 1 = 6 C ; .\ C = — |-. 

Examples. 

1. Find n*- 1 )* 8 -. ^ s . log-^±ili c . 

J a; 2 + 6a; + 8 " (aj + 2)# 

*J X —\~ X — u 



)da? . t /a;— 2\* . ~ 



4 ° ^ + 2 

174. Case II. When some of the simple factors of the de- 
nominator are real and equal, to every set of equal factors, as 
(x — a) 11 , there corresponds a series of n partial fractions of the 

, A . B . L 

form — — + — -^ + ••• 4 



(x-a) n (x-a) 11 - 1 x-a 

dx 



/d 
z r. 
(x—1)' 



Assume T7 — ,N2/- ■ -.x = 7~ — ~9 + — : — — + -. ■ , • 0) 



' 2 (a+l) 
1 ^ 1 ^ 1 

(a,-l) 2 (a, + l) (a;-l) 2 x-1 »+l 

Clearing (1) of fractions, we have 

1 = .4 (x + 1) + J5(^ - 1) + C (x - l) 2 
= (B + C)x 2 + (A-2C)x + A-B + C. (2) 

Equating the coefficients of like powers of a?, we obtain 

£+C=0, .4-20 = 0, and ^-£ + (7 = 1. (3) 
Whence A = £, C=i, and J5 = — J. (4) 



CASE II. 



191 



Substituting these values in (1), and integrating, we have 

/ dx 

J2(x-1) 2 J ±(x-l) J 4(0^+1) {) 



x+l 



1 



+ c. 



og (z-iy 2{x-i) 

Equation (5) is true ; for the values of A, JB, and C given 
in (4) satisfy (3), and hence make (2), and therefore (1), an 
identical equation. 

Examples. 

, va C (2x — 5)dx 7 .11, 35+1 . ^ 

L Fmd J (g + 8)( g +i)' - i^+T) + T l0g ^T3 + a 



+ log(a;+l) + C. 



+ 8o; + 4 



J x 3 -\- 5 x 

3 r(2x*+7a?+Gx+2)dx . j 
'J a* 4 + 3^ + 2^ 8 



# + 2 



x(x + l) 



x \i" 



^+27 



-±+0. 

a? 



175. When the simple factors of its denominator are imagi- 
nary, a fraction may then also be decomposed into partial 
fractions of the forms in Cases I. and II. ; but, since the inte- 
grals obtained from these would involve the logarithms of 
imaginaries, we seek other forms for the partial fractions. 

Since the imaginary roots of an equation always occur in 
conjugate pairs, the imaginary factors of the denominator will 
occur in pairs whose products are real quadratic factors of the 
form {x — c/) 2 + 6 2 . Hence, when its simple factors are im- 
aginary, the denominator can be resolved into real quadratic 
factors. 

Case III. When some of the simple factors of the denomina- 
tor are imaginary and unequal, to every factor of the form 
(x — a) 2 -f-b 2 there corresponds a partial fraction of the form 

Ax + B 
(x-a) 2 -fb 2 * 



192 INTEGRATION OF RATIONAL FRACTIONS. 



Let it be required to find | 

J aj* + x 2 - 2 

Assume * = -JL + -*_ + «*±£ (1) 

(x+l){x-l)(a?+2) x-\-l x-l x 2 -\-2 KJ 

Clearing (1) of fractions, we obtain 

x 2 = A(x - 1) (x 2 + 2) + B(x + 1) (a 2 + 2) 
+ (G* + I>)(ar -1) 
= (^L + B + <7)ar 3 -{- (5 - .4 + D)a 2 

+ (2^1 + 2£-0)x + 25-2^4-1). (2) 

Equating coefficients of like powers of a?, we have 
A + B+C = 0, B-A + D = l, 
2A + 2B-C-0, 2B-2A-D = 0. 



(3) 



Whence A = -±, -B = i, (7=0, 7)=f. (4) 

Hence f /f =-1 f_^L + i fj?£L + 2 fd*_ (5) 
J a;4 +iC 2_2 QJ x+l 6J x-l 3J x*+2 K J 

= ilog — + - V2 tan^ 1 — + O. 
6 & a + 1^3 V2 

Equation (5) is true ; for the values of A, B, (7, and D given 
in (4) satisfy (3), and hence make (2), and therefore (1), an 
identical equation. 

Examples. 



1. Find f 1^- -• 

J (a-f-l)(o; 2 + l) 



Ans. itan- 1 ^ + ilog ( a ^ 2 + 1 ) i + Q. 

2 ^2 ° x + 1 ^ 



C x*dx m j (a? + 2)c 

' J aJ* + 3a^ + 2 s (x 2 -f-l)^ 

J (aj-l^ + l) 20-1) 4 & a 2 -fl 



/ ardx 



CASE IV. 193 



4 *l-x 2 



J a 3 + a 2 + « + 1 

Jlog (a? + 1) +ilog(^ 2 + 1) - itaa-'x + 0. 

G. f-*L. 
Ja^ + 1 

r ja; _ 1 f_efo_ _ 1 / "(a; — 2) eta ; . 
J a; 3 + 1 "" 3 J x + 1 3 J a? - a? + 1 .' 

r (x — 2)dx _l r(2x—l)dx 1 /" 3 da; 
J a^_ ^ J. i - 2 J ar-x + 1 2Jx 2 — x + l' 

A 3 + l 6 & a; 2 -a;+l V3 V3 



/ <7a? 
1-5 



1, a?-fajr|-l . 1 , _i2a; + l , ~ 

-loo-— — I ! tan 1 ■ h C. 

6 & ^_2a; + l V3 V3 



176. Case IV. IFAew some of the simple factors of the denom- 
inator are imaginary and equal, to every set of equal quadratic 
factors of the form [(x — a) 2 + lr] n there corresponds a series of 
n partial fractions of the form 

Ax + B Cx + D Lx+M 

[ (x - a) 2 + b 2 ] 11 [(x - a) 2 + b 2 ]*- 1 (x - a) 2 + b 2 ° 

In any example under this case, by clearing the assumed 
equation of fractions, and equating the coefficients of like pow- 
ers of x, we should, as in the first three cases, evident!}- obtain 
as many simple equations as there are indeterminate quantities ; 
and the values of A, jB, (7, etc., determined by these equations 
would make the assumed equation an identical one. 

It remains to be proved that we can always integrate a frac- 

.. - , T - (Ax-\- B)dx 
tion of the form — * ' ; _ • 



194 INTEGRATION OF RATIONAL FRACTIONS. 

Let z = x — a ; then x = z + «> da; = dz, and 

J [(a;-a) 2 + 6 2 ] w ~J (z 2 + & 2 ) w 

_ f Azdz r(Aa+B)dz 

J (z 2 + b 2 ) n -J (z 2 + b 2 ) n 

_ — A r(Aa + B)dz 

2{n-l){z 2 -\-b 2 ) n ~ l J (z 2 + b 2 ) n 

The method of finding the integral of the last term is given 
in § 185. 

Examples. 

r(a? + x-l)dx m 
J (a; 2 + 2) 2 

Arts. ilog(a; 2 +2) + -? f o da? • 

Assume ^ + x ~ l = Ax + i? + £*±£ 

(a^ + 2) 2 (x 2 + 2) 2 ^ a; 2 + 2 

2# p (s»-s+l)da logi^LDl-jtan-^+a 

J ajS + ^ + aj+l B (V + 1)* 

3. f _^ tog O- 1 ) 1 L_ tan- 1 — + C. 

J a?- x * + 2 x -2 *(x 2 + 2)h 3 V2 V2 

J (a;+2) 3 (a; + 2) 2 a;+2 ^ ^ ; ^ 



CHAPTER XIV. 

INTEGRATION BY RATIONALIZATION. 

177. It has been shown that any rational differential is inte- 
grate ; hence an irrational differential which does not belong 
to a known form can be integrated, if we can rationalize it ; 
that is, if we can find its equivalent rational differential in terms 
of a new variable which is some definite function of the given 
variable. 

178. A differential containing no surd but those of the form x d 
lan be rationalized by assuming x = z n , in ivhich n is the least 
common multiple of all the denominators of the several fractional 
exponents o/x. 

For, if x = z n , the values of x, dx, and each of the surds, will 
be rational in terms of z\ hence the function of z obtained by 
substituting these values in the given function will be rational. 

Examples. 
1. Find C xh "f" dx. Ans. |^-Aoi + <7. 

J 2x1 8 3 -r 

Assume x = z 6 ; then 

x* = z 3 , xi = z 4 , xi = z, and dx = 6 z?dz ; 
.'.C xl ~f dx = Ctzit^dz = 3 f( z 7 _ z ^ dz 

2 * f^Zlh' !^-i^ + |^-f^ + Alog(l+2^)4C. 
•/ x* -\- zx* 



i96 INTEGRATION BY RATIONALIZATION. 

179. A differential containing no surd except a + bx affected 
with fractional exponents can be rationalized by assuming a + bx 
= z n , in which n is the least common multiple of the denominators 
of the several fractional exponents. 

For, if a -f bx = z n , the values of x, dx, and each of the surds, 
will be rational in terms of z. 



Examples. 

1. Find C- -^ -. Ana. 2tan- 1 (l+xH + C. 

Assume 1 + x — ?} ; then 

(l + x)i=z\ 

(l-f-#)2 = 2, and dx = 2zdz; 

. r dx C 2zdz _ 2 f_dz__ 

"J (l + x)i + (l + x)l J z *+z~ J z 2 +l 

= 2tan~ 1 z+ 0= 2tan- 1 (l-f-«)2 + (7. 

dx 2 (2 a + to) „ 

&»)* bWa + bx 



2. f-*< 

J (a + 



- C dx It Va + to — Va . t 

5. I • — -log — — | = +logc. 

J x Va -f- bx Va Va+~to -f- Va 



+ a) 2 1-x 

7 2 



180. ^4 differential containing no surd except Va + bx -+■ x 2 
can be rationalized by assuming Va -f- bx -f- x 2 = z — x. 



TRINOMIAL iSUBDS. 197 

Let Va + bx -f- x 2 = z — x; 

then a-\-bx = z 2 — 2 zx, 



z 2 — a 
x = 



and 



b + 2z 

j _ 2 (z 2 +jg-j- a) dz 

**" (& + 2z) 2 ~~' 

/ — r^ — ; — ■> r n z 2 -\-bz-\-a 
Va -f 6ic + xr [ = z — x] = — ! ! — 



Hence, as the values of x, dx, and V« -\-bx-\-x 2 expressed in 
terms of z are rational, the given differential when expressed in 
terms of z must be rational. 



181. A differential containing no surd except Va + bx — x- 
can be rationalized by assuming 

Va + bx-x 2 [ = V(x-/3)(y-x)] = (x-/?)z, 
in which ft and y are the roots of x 2 — bx — a = 0. 



Let -y/a + bx — oj 2 [= V(» — ft) (y — *)]= (x — /3)z 

then y — x = (x — ft) z 2 , 

x^4±y, 

z 2 + l ' 

(z 2 + l) 2 

and Va-h&aJ-aj 2 [= (a? - /3)»] = ^~^ g . 

z 2 -f- 1 



Hence, as the values of a?, da?, and V« + &# — &~ expressed in 
terms of z are rational, the differential when expressed in terms 
of z must be rational. 

Examples. 

1. Find f dx Ans. log[(2aj+l+2Vl + a? + «")c]. 

J Vl + a + cc 2 



198 INTEGRATION BY RATIONALIZATION. 

Assume Vl -\-x-\-x 2 = z — x; 

2-2 1 

then x = - -, 

2z + l' 

(2z + iy ' 

and Vi +fl ; + ~^r = z_ar| = g2 + g + 1 . 

L J 22 + 1 

.*. P dx = f_!^-=log[(23 + l)c] 

^ Vl+aj + ar 2 J 2. + 1 bLV T ; J 



/: 



= log [(2 a; + 1 + 2 VI + x -j- x 2 )c\ 
dx 



V 2 — x — x 2 
The roots of x 2 -f- x — 2 = are — 2 and 1 ; 
.-. 2 - x - x 2 = (a? + 2) (1 - a?) . 



Assume V2 — a; — x 2 [= V(a; + 2) (1— a?)] = (a? + 2)z ; 
then 1 — x = (x + 2) z 2 , 

l-2z 2 



a? = 



dx = 



z 2 + l 7 

— 6zdz 

(^ 2 + I) 2 ' 

3z 



and V2 — a? — a? 2 



z 2 + l 
J -W2—X-X 2 J * + 1 



V2 — x — a? 2 

=3 coir 1 / 

,x + 2 



2 cot' 1 f^—^Y 



3. f ^ 2cot- 1 ^Y + C. 

J Voa;-6-aj 2 V*" 2 / 



4 rV2^+^ # iog( a j+l + V2a?+aj 2 ) ± 4- C. 

J x 2 ^ + V2a?-j-a; 2 



BINOMIAL DIFFERENTIALS. 199 

182. Binomial Differentials. Differentials of the form 

x m (a + bx n ) p dx, 

in which ra, n, and p represent any numbers, are called binomial 
differentials. When p is a whole number, the binomial factor 
can be expanded, and the differential exactly integrated by pre- 
vious methods. In what follows, p is regarded as fractional; 

and, in the next section, we will represent it by -, r and s being 
whole numbers. 

r 

183. Conditions of Rationalization of x m (a -f- bx n )* dx. 

I. Assume a -f- bx 11 = z s ; 

then (a + ox n ) ~* = z r ; ( 1 ) 

1 m 

(z* — a\n fz s — a\n , . 

" = irn ' x {—) ' (2) 

1-1 

and efcc= — 2 S_1 [ — H— ] dz. (3) 

6>i V h J 

Multiplying (1), (2), and (3) together, we obtain 

m+l 

x m (a + &af)Tda; = JL z r+s-if^zSL\ n dz. (4) 

bn \ b J 

The second member of (4) is rational, and therefore integra- 
te, when m ' is a whole number or zero. 
n 

II. Assume a -f- foc n = z s # n ; 

then a; n = a(2 s -6)- 1 ; (1) 

1 1 wi m 

# = a»(z s — &)"*, x m = an(z 8 — 6)"» ; (2) 

and ^ = --«^ s - 1 (^- & )~^ 1 ^- ( 3 ) 



Multiplying (1) by 6, and adding a, we obtain 

a + bx n = ; 

z 8 — b 



200 INTEGRATION BY RATIONALIZATION. 

.-. (a + bx n )T = aJ(z s -by^z r . (4) 

Multiplying (2), (3), and (4) together, we have 

x m (a + bx n ydx 

,=: - -a~ + s (z s -b)~ { ~ + ^ +l } z r+s ~ l dz. (5) 



The second member of (5) is rational and integrable, when 

771 I I V 

— — — f- - is a whole number or zero. 

n s 

Hence, x m (a -f- bx n )^dx can be integrated by rationalization. 

I. When — ±— is a whole number or zero, by assuming 

a -\- bx n = z s . 

n% 1 1 t 

II. When — — — (__ is a whole number or zero, by assuming 

n s 

a -\-bx n = z s x n . 

Examples. 
1. Find fx 3 (a + bx 2 )-hdx. Ans. h% * ~ 2 a (a + bx 2 ) \ + C. 

•J o 

t)\ I 1 

Here — — is a whole number, and s = 2 ; hence we assume 

a -j- bx 2 = 2 2 ; 
.\ (a + bx'yh^z- 1 ; (1) 

a =(t=*)\ * = (*=*)*; (2) 



wi -4- 1 wi -4- 1 t 
* When 1 is a negative integer, or -\ 1-1 is a positive 

integer, the exponent of z s — b being negative, the given differential will be 
reduced to a rational fraction whose integral may be obtained by the 
method of Chapter XIII. But, as this method usually gives a complicated 
result, it is generally expedient in such cases to integrate by using the 
formulas of reduction given in § 185. 



EXAMPLES. 201 

and dx=-(— Ydz. (3) 

b \z L — a J 

Multiplying (1), (2), and (3) together, we obtain 
Cx*(a + brf)-ldx = \ C(z 2 - a)dz 

2 C—^^— 4 — 3 ar 2 , ^ 

"J(2-3a^)§" 9(2-3^)5 

3. r^(i + ^-i^. (2^-i)d+^i +c . 

•/ 3 a; 3 

Here — — — f- - is a whole number, and s = 2 ; hence we 
n s 

assume 

l+z 2 = zV; .-. a? = (z 2 -l)-\ 

x = (?-l)-h, ar^O^-l)", (1) 

cfa = -(> 2 -l)-izcfe; (2) 

and (l + ^)-| = A + _L_Y l = ^(^_i)^ (3) 

Multiplying (1), (2), and (3) together, we have 

f da; = - f(^-l)^ 
J a; 4 (l + ar 9 )i J v ; 

-* T + c ~ 3^ +c * 



:. I — — - • ^log — "" \-\OQ-C. 



1 

! )* a ° Va 2 + a 2 + « 



Ti Ki u " lQ g / o ; hlo « c - 

a; ( cr — ar ) 2 a ,/„2 _ ~2 , „ 



(« 2 -^) 5 a Va 2 - a? + a 

aa; 



J agg ax ^ 

(l + aj 2 )!* 



202 INTEGRATION BY RATIONALIZATION. 

Miscellaneous Examples. 
l.'Find f(lzjA±dx . 

J 1 —X3 

Ans. 6|>ajS + \xi-\xl + \xl - ±xk + xl- log(l + a#)] + 0. 
2. f S J-loa V ^ + Vl + J? + W 

3 - f^ tt; rr 2tan- 1 (l + ^)^ + C. 

4. J(a + &*)**<** 2(a + to)l/a^to_a\ + 0> 

J (a+bx) 1 * b 2 v J \ 5 2y 

!. f ^ - . J- tan-^Mz^Y + a. 

+ 0. 



. r— 

^ (i+ 



(6a;-a)i Va V a 

dx 2 /2 — a>\* 






dx ion( a+VT+^+# 



(1 + a) VI + a + ic 2 \2 + a? + VI + x + a? 2 
9. f^a + b^idx. ( a + 6ffi2)8 ( 56 3 5 y 2q ) +a 

in f ^ ^(20? + 3) c 

' J (1+^)1* 3(1 + ^)* 

1 1 /^ * v?dx x s ^ 

J (a + 6aj«)l" 3a(a + 6^)i 

1 9 f * dx a-\- 'Ibx 2 p 

' J : ${a + bx*)£ a 2 x(a + bx 2 )^ ' 



CHAPTER XV. 

INTEGRATION BY PARTS AND BY SERIES. 

184. If u and v be any functions of x, we have 

d(uv) = udv -f vdu. 
Integrating and transposing, we have 

I udv = uv — I vdu. (1) 

Equation (1) is the formula for integration by parts. It 
reduces the integration of udv to that of vdu ; and, by its appli- 
cation, many useful formulas of reduction are obtained. 

Examples. 

1. Find I x log xdx. Ans. ^x^logx — ^x 2 -{- C. 
Assume u = log£, and dv = xdx ; 

then du = — , and v = — 

x 2 

Substituting these values in formula (1), we have 

| x log xdx = ^x 2 log x — J \xdx 

= ix 2 \ogx-ix 2 +C. 

In each example, the values of u and dv must be so assumed 
that vdu is a known form, or nearer one than udv. 

2. I log xdx. x(logx — l) + C. 

3. fxe^dx. er{- - -V) + G. 
J \a a 1 ) 

Assume dv — e ax dx. 



204 INTEGRATION BY PARTS. 

4 



Cx n log x dx. -^— - Aog x — \-i-C. 

5. I sm^xdx. a?sin _1 # + (l — x 2 )h-\- C. 

6. J tan -1 x dx. #tan _1 # — log^ + as 2 )* -f (7. 

7. faf(a - a; 2 )^. -i^(a - a 2 )« -^(a- x 2 )i + (7. 
Assume ?t = x 2 . 

8. J cc cos xdx. x sin a -f- cos as -f- (7. 

185. Formulas of Reduction. We will next apply the formula 
for integration by parts to the binomial differential, 

x m (a + bx n ) p dx* 

in which p is any fraction, but m and n are whole numbers, and 
n is positive. 

I. Let dv = (a + baf t yaf^ i dx, and w = a m - ,l+1 ; 
then v = ± — — - — , and du = (m — n 4- l)x m ~ n dx. 



Hence, by the formula, we have 

nb(p-\- 1) 



J>(« + b*y dx = ^-' +1 (» + ^r 1 



m — n 



nb(p 



-\-l)J 



* That any binomial differential can be reduced to one of this form is 
evident. For, let the given differential be (.r 5 — x 5 ) 5 x~*.dx , then, by mul- 
tiplying the first factor and dividing the second by (r)s, we obtain 
(1— x*)%x~*dx. Putting x = z Q , we have 6 (1 — z b )% dz, which is of the form 
required. 

The formulas of reduction are true for fractional and negative values 
of m and n, but they are not generally useful in leading to known forms, 
unless m and n are whole numbers, and n positive. 



FORMULAS OF REDUCTION. 205 

Now Cx m ~ n (a + bx n ) p+1 dx = Cr i ~ n {a -f bx H ) p (a + bx n )dx 

= a Cx m -' l (a -f bx H ) p dx + b Cx m (a + bx n ydx. 

Substituting this value in (1), we have 

Cx m (a + bx n Ydx = s m ~ w+1 (« + fc« H ) J,+1 
J v y w6(jp+l) 

nb(p+l) J 

- m ~ n + 1 f^(a + bx n Ydx. 

Transposing the last term to the first member, and solving 
for I x m (a -\- bx n ) p dx, we obtain 

£ m (a + bx n ) p dx = * — ! i — 

b(np + m+l) 

- a ( m - n + 1 \ f x ™-n( a + 6aJ»)'da?. (A) 
6(wp + m + l)J v y v 

By formula (A) the integration of x m (a + bx n ) p dx is made to 
depend upon that of another differential of the same form, in 
which m is diminished by n. By a repetition of its use, m may 
be diminished by any multiple of n. 

Formula (A) evidently fails when np + m + l = ; but in 

that case + p == ; hence the method of integration by 

rationalization (§183) is applicable, and the formula is not 
needed. 

II. It is evident that 

fx m (a + bx ll ) p dx = Cx m (a -f bx n ) p ~ 1 (a + bx ll )dx 

= aCx m (a + bx n y- 1 dx 

+ b Cx m+n (a + bx n y- x dx. (2) 



/■ 



206 INTEGRATION BY PARTS. 

Applying formula (A) to the last term of (2) , by substituting 
in the formula m + n for m, and p —1 for p, we obtain 

b fx m+n (a + bar)'- 1 */} = xm+ \ a + hxn Y 
J np -f- m + 1 

np-f- m + U 
Substituting this in (2) , and uniting similar terms, we have 

V(a + 6af)*daj = a*"* 1 (<* + &*")* 
wp + m -+- 1 

H ^ faf , (a v +".6afy^ 1 daB. (B) 

np + m+U 

Each application of formula (B) diminishes p, the exponent 
of the binomial, by unity. It fails in the same case that (A) 
does. 

III. When m and p are negative, we need formulas to increase 
rather than to diminish them. To obtain these, we reverse for- 
mulas (A) and (B) . 

Solving (A) for I af l-w (a -f- bx n ) p dx, and substituting m + n 

for m, we obtain 

x m+1 (a-\-bx n Y +1 



/x m (a + bx n ) p dx 
v J a(m+l) 

_ 6(np+n+m+l) f^+^+^w*. (C) 

a(m + l) J V ; V ; 

Formula (C) enables us to increase m by n at each applica- 
tion. It fails when m -f 1 = ; but in that case the differential 
can be rationalized (§ 183). 

IV. Solving (B) for I x m (a-j- bx n ) p ~ 1 dx, and substituting p-f-1 
forp, we obtain 

I # m ( a 4- &sc n r aa? = — ! — ■ 

J V J an(p + l) 

np + n + m+1 j^ ( ^), + i^. (D) 



EXAMPLES. 207 

Formula (D) enables us to increase p by unity at each appli- 
cation . 

The mode of applying formulas (A) , (B) , (C) , and (D) will 
be illustrated by a few examples. 



Examples. 



1. Find f f dX ■ 
J (a 2 — ar)5 



(a 2 -ar°)* 

Here m = 4, n = 2, p — — |-, a — a 2 , and b = — 1 . Hence, by 
applying formula (A) twice in succession, this integral will evi- 
dently be made to depend upon 

dx 



f 



V a 2 — x 2 



Substituting these values of m, n, etc., in formula (A), we 
obtain 

Cx\a 2 - x 2 )-ldx = - ix?(a 2 - x>)k 
+ %a 2 ftf^-tfyhdx. (1) 

In like manner, we obtain 

I ^{a 2 — xPJ-zdx^ — %x(a 2 — a?)h 
+ i« 2 f(a 2 -^)-Mx. (2) 

Now, C(a 2 -x 2 )-hdx= f f dx =sm- 1 -+C. (3) 
From (1), (2), and (3), we obtain 

f a * i<to = - (ioj 8 +| a 2 x*) Va 2 -^ 
•^ Va 2 — ft 2 

4- 1 a 4 sin- 1 - +0. 

a C x 2 dx A x /— o i a 2 • i# . >-, 

2. I — • -4ns. V«" — or H sin -1 - + C. 

J Va 2 - x 2 2 2 a 



208 INTEGRATION BY PARTS. 

3< r tfdx m _^(^ H .2a 2 )(a 2 -^ 2 )^+C. 

•^ Va 2 — x 2 

— ' . is made to depend on what known 

Va 2 — x 2 
form when m is positive and even ? On what known form when 
m is positive and odd? 

4. f ^ dx ■ -V?T?-ia 2 log(aj+V?+^) + C. 

5> | " g 8 ^ . i( aj 2_2a 2 )V^4 r 5+C'. 

/* x m dx 
By formula (A) , I — ' is made to depend on what known 

J Va 2 -{-£C 2 
form when m is positive and odd ? On what known form when 
m is positive and even? 

Jvr^ 5 v.5 5.3 5.3; 

7. f-4 

*^ Va 2 



; cte 



a? 5 . 5aV . 5-3aV\ /-» \ . 5 -3a 6 . _,« . „ 

1 (- V<r — ar -\ sin - + C. 



x 2 

6 + 6-4 ' 6-4-2J ' 6-4-2 a 

8. f(a 2 -x 2 )hdx. ix(a 2 -x 2 )h + ±a 2 sm~ 1 -+C. 

«y CI 

Apply formula (B) once. 

9. Cx 2 (l-x 2 )hdx. ix(2x 2 -l)(l-x 2 )h^.^sm- 1 x-hC. 

10. Cx\l-x 2 )ldx. 



-- C dx Va 2 -r p 

' Jrf{a*-a?)h a 2 x + 



EXAMPLES. 209 



12. f *? - Vfl -^ + J -log - + 0. 

For f da; see § 183, Ex. 5. 
J x~\/a 2 — x 2 

, 8 . Atf-iflMy . al y +(tt ,_^ )t + g , 

J 2/ a + (a 2 -r) 5 

Apply formula (B) . 

14. f(a 2 +x*)*dx. ixVtf^ 2 +ianog(x+Va?+tf)+C. 

15. f— ^ — * + 1 feur^ + a 

J (c^ + ar*) 2 2a 2 (a 2 + 2a 3 a 

Apply formula (D). 

16. f(l—sj)*<kr. ia;(l-iB 2 )i+ta;(l-iB 2 )5+fsin- 1 a; + 0. 

17. f ^ . -(2aaj-ic 2 )l + avers- 1 -+0. 

*^ V2a«-.T 2 a 

. /» #da; /» j 

Write I , = iu the form | x*(2a — x)~*dx, and apply 

J v2ax — x~ J 

formula (A) once. 

18> J " a 2 da . _ x±3a ^/2ax-x 2 + fevers" 1 - + O. 

J -V2CIX-X 2 2 a 



-. q U/ C&37 



V 2 aa; — ft 2 

! -J — ! L V2 a# — x 2 -f fa 3 vers -1 - -f- C. 



6 
20 



r dx 

J (cf + x 2 ) 



x Sx . 3 , ,£C , ^ 



4a 2 (a 2 + ^) 2 Sa^^ + aJ 2 ) 8a 5 a 



• f **** . x(S-x*) , 



21, 

(1-z 2 )* 2(l-ar 9 )£ 



210 INTEGRATION BY PARTS. 

186. To integrate the logarithmic differential </>(x) (7o#x) n dx, 
in which <£(x) is an algebraic function, and n is a positive ivhole 
number. 

Let x 2 (\ogx) 2 dx be the function. 

Assume dv = x 2 dx, and u = (logx) 2 ; 

then dw=21oga; — , and v = ^x 3 . 

x 

Hence, by the formula for integration by parts, we have 

I x 2 (logx) 2 dx = ^x s (logx) 2 — J J x 2 logxdx. (1) 

Applying the formula to the last term of (1), we have 

I x 2 logxdx = \x 3 log a? — I ^x 2 dx. (2) 

From (1) and (2), we obtain 

Ca?(logx) a dx = -Jar 3 [(logo?) 2 - f loga; + -§] + C. 

Hence, to integrate <jt(x) (logx) n dx, we assume dv — <j>(x)dx. 
and, by successive applications of the formula for integration 
by parts, reduce the exponent of log a? to zero. In this way 
the integration of the logarithmic differential is reduced to the 
integration of algebraic differentials. 



Examples. 

dx x 



1. Find J x s (\ogx) 2 dx. Ans. i a; 4 [(log a;) 2 — -J- logo; + -J] + 

o r logxdx X 

J (l-M) 2 ' 1+x 



log a; — log(l + x) + C. 



3. JV ( iog^. gi[(iog») 2 -4T log * + (^ir] +a 

4 _ J(M! da; . __i_ [(loga;)2 + |loga;+ | ] + o. 



EXPONENTIAL DIFFERENTIALS. 211 

187. To integrate the exponential differential, x n a mx dx, when 
n is a positive integer. 

Assume civ = a^dx, and u = x n ; 

then du = nx n ~ 1 dx, and v = 



mloga 
Hence, by the formula for integration by parts, we have 

x n a nx dx = -^ — x'^ar^dx. 

mloga mloga J 

By successive applications of this formula, the exponent of x 
is reduced to zero, and the proposed integral is made to depend 
upon the known form 



/• 



a^dx. 

Examples. 
1. Find Cx-e^dx. 

Assume dv = e^dx, and u = x 2 ; 



then | or e ax dx = ~e ax or \xe^ dx. 

J - a aj 

Again, J x e°* dx = - e ax x — - | e™ dx. 

/e ax f 2x 2\ 
x>e ax dx = — [x 2 - — + — + C- 
a \ a a- J 

^ C % ^1 e^/o 3 2 . 3-2 3-2.1\ , n 

2. I aPe^dx. — I x 3 xr -\ x — - ] + C. 

J a \ a or a° J 

3. lx 3 a x dx. 



loo- a 



o 3-x 2 . 3-2 -x 3-2-1 

xr — < h 



loga (log a) 2 (loga) 3 _ 



4. Write out the integral of e x x 4 dx, according to the law 
of the integrals in Examples 2 and 3. 

188. To integrate the trigonometric differential sin m x. cos n xdx. 
Let sin x = z; 

then sin m x = z n \ cos' 1 x = (1 — z 2 ) 2", 



212 INTEGRATION BY PARTS. 

and dx = (l— z 2 )~^dz. 

sin m £ccos M icdic= j z m {\ — z 2 ) 2 dz. (1) 

Or, letting cos x == 2, we obtain 

sin w a;cos M fl?^= I — z w (l — z 2 )~2~dz. (2) 

Hence, whenever the binomial differential in (1) or (2) can 
be integrated, the given differential can be. 

This method of integrating sin m xcos n xdx is used in those 
cases to which the shorter methods of § 63 are not applicable. 



Examples. 
1. Find j sm 6 xdx. 

Put sin x = z\ 

then dx = ( 1 — z 2 ) ~i dz, 

and lsm 6 xdx = I z G (] — z 2 )~* 



dz 



6 6-4 e-4.2/ v } 

-5^-sin- 1 ^ + C § 185, Ex. 7. 

6-4-2 



COS Xf . K , r • Q ,5«3. \ 

snrsc + 4sura5 -\ smo? 

4 4-2 J 

x+C. 



5-3 
5.4.2 



2. i $m 4 xdx. Ans. — J cos a; (sin 3 #-f-f since) -ff x + (7. 

3. fcos 4 cccte. isin:r(cos 3 £ + f cosaj)+fa?+ (7. 

4. I sin 2 aJCOS 2 icc?a?. ^sin 3 £ccos£c — -§-sina;cosaj + -|-a;+ (7, 

or \(x — i sin 4 #) + C. 
Let sina? = :<;, and, for j z 2 (1 — z 2 )*dz, see §185, Ex. 9. 



TRIGONOMETRIC DIFFERENTIALS. 213 

The second form of the integral is obtained from the first by 
use of the relations, 2 sin x cos x = sin 2 x, and 2sin 2 &=l — cos 2x. 
By a similar transformation, any differential or integral expressed 
in powers of since and cos a; may be found in terms of the sines 
and cosines of multiples of x. 

C • 2 4 7 sin#/cos 3 # cos 5 a? . cosaA , x . n 

o. I sin 2 £cos 4 #cfa\ [ H \-C. 

J 2 V 12 3 8 J T 16 T 

6 . C. ^L_. — l + log tan x + C. 

J since cos 3 a? 2 cos- a; 



r r dx 

J sin 3 a 



^L + ^logtanf+C. 
2 sin 2 a? 2 



• -i-cos^a? + cos a; + I02; tan- -f- C. 

x 3 8 2 

189. To integrate x n sm(ax)dx, cmcZ x n cos(ax)dx. 
Assume % = a?*, and apply the formula for integration by 

parts. Each application of the formula will evidently diminish 
n by unity ; hence, when n is a positive integer, the integral 
can be made to depend on the known form 

J cos (ax) dx or j sin (ax) dx. 

190. To integrate e ax sm n xdx, and e ax cos n xdx. 
Put dv = e ax dx, and u = sin 71 a? ; 



then 


v = -e™, 
a 


and 


du = n sin n_1 fl7 cos a; da?. 




.*. ( e™ sin' 1 xdx = - e ax sin n x 
J a 




I e ax sm n ~ 1 XQ,o&xdx. 

aJ 


Again, 


put dv = e^dx, and u = sin' l ~ 1 a?cosa. ; 


then 


v = -e ax , 
a 



(1) 



214 INTEGRATION BY PARTS. 

and du = (n — 1) sin n ~ 2 # cos 2 xdx — sm n xdx 

— (n — 1) sin n ~ 2 xdx — n $m n xdx. 

[Since cos 2 a; = 1 — sin 2 ct\] 

J e ax s [ n n-i x Q sxdx — -e ax sin"" 1 ^ cos a? 
a 



n — 1 



| e ax sin n ~ 2 xdx _l. - J e ax sm n xdx. 



Substituting this result in (1), and solving for j e ax sm. n xdx, 
we obtain, 

> mn*xdx = ^Bin^^Bing-ncosa?) 

?i 2 -f-a 2 

w fo~ V fe ax sin w - 2 x^. (2) 

?r + ar J 



s< 



By repeating this process or the application of this formula, 
n is reduced to zero or unity ; and the integral is made to de- 
pend upon the known form | e ax dx or the form j e ax sinxdx. 
The value of the latter form is obtained directly from (2) by 
making n = 1 . 

In like manner I e ax cos n xdx can be obtained. 

Examples. 

1. Find IxPcosxdx. 

Here u — x 2 , dv = cos x dx, 

v = since, and du = 2xdx. 

.*. I x 2 cosxdx = x 2 sinx — 2 I x sinxdx 

= « 2 sin» + 2^cos^— 2 I cosxdx 

= x 2 sinx-f- 2x cosa? — 2 since -f- O. 

2. IxPsinxdx. —afcosx+Sx 2 sinx-}-6x cosx—6smx-\-C. 



TRIGONOMETRIC DIFFERENTIALS . 215 

/pax 
e ax sinxdx. — (a sin a; — coso?) + C. 

a? + l 

e x sm 3 xdx. — (sin 3 ^ + 3cos 3 aj + 3sin^— 6cos#) +C. 

X.\J 

dx 



191. To integrate 



a -\-b cosx 

dx 



r dx _ r 

J a+b cosx- J a ( cQs2 x 

-/■ 



+sin 2 - )-\-b cos 2 sin 2 - 

V 2 27 \ '2 2 



-/ 



(a + 6) cos 2 - +(a-6) sin 2 - 



sec 2 -c?cc 



(a + 6) + (a - 6) tan 2 - 



■/■ 



tan | 



(a + 5) + (a -6) tan 2 - 



which is readily reduced to the known form, 

/dx C dx 

7+tf or J ^w' aceordiug as a > or < 6> 



In like manner f can be found. 

J a 



dx 
-f- b sin # 



192. To integrate the anti-trigonometric differentials, 

f (x) sm -1 xdx, f(x)cos _1 xdx, f (x) fcm^xdx, etc., 

in which f (x) is an algebraic function. 

Assume dv =f(x) dx, and apply the formula for integrating 
by parts. One application of the formula will evidently make 
the integral depend on an algebraic form. 



216 INTEGRATION BY SERIES. 



Examples. 

1. Find f^^-tan- 1 ^ 
J 1+x 2 

Qu dx dx 

Here dv = — '— - — dx , u= tan -1 #, 

1+x 2 1+x 2 

v = x — tan" 1 ^, and du = - •• 

1+x 2 



-tan- 1 ^ = x tan _1 # — (tan -1 ic) 2 

1+x 2 v ' 



/ xdx r t&rr^xdx 
l+x 2 J l+x 2 
= x tan" 1 ^ — (tan- 1 ^) 2 

- ilog (1 + a?) + i (tan-^) 2 + G 



= tan" 1 ^ {x — -J tan -1 a;) — log Vl +X 2 + C. 

2. I xcos^xdx. ^x 2 qos~ 1 x — ix{l — x 2 )* + ±sm~ 1 x+ 0. 

3. Cafsm^xdx. i^sm" 1 :*; +±(x 2 + 2) Vl^x 2 + C. 

193. Integration by Series. When we cannot, by any of the 
preceding methods, integrate a given differential exactly ; or, 
when the integral obtained by them is of a complicated form, 
we can develop the given differential in a series, and integrate 
its terms separately. Moreover, integration by series furnishes 
a simple method of developing a function, when we know the 
development of its derivative. For examples of this method of 
developing functions, see §§ 107, 108. 

Examples. 
1. Find Cx*{l-x 2 )ldx. 

(l-x^l^l-^x 2 - ix* - ^ — ; 



EXAMPLES. 217 

.-. Cc£(l-a?)kdx= Cxh(l-^-ia^-^x e )dx 

which is the required integral for x < 1 and > — 1 . 

2. jx*(l-a?)ldx. i^_^^_J^^__ i _^_... + a 

3. Prove that log(a+x) = loga + - - ^ + ^-_^L + ..., 

a 2cr 3a° 4 a 4 

dx 

by first integrating directly, and then by series. 

a -\-x 

dx 



4. Develop log (a + Vl +X 2 ) by integrating 



Vl^M 



r dx 

J (l+x* 



a i / , /T-T—o\ I* 3 , l-3af 1.3.5a 7 , 

Ans. log(#+ Vl4-ar) = # — 

SV ' 2 3 2-4 5 2-4-6 7 

5. Prove that 

1 a 5 , 1 • 3 a 9 1 • 3 - 5 x 13 . . n 

(l+x*)h 2 5 2-4 9 2-4-613 

194. While we can differentiate any given integral, we can 
exactly integrate but a small number of differentials. One 
great reason for this seeming difference in the perfection of the 
two branches of the Calculus is that the integral is often a 
higher or more complex function than its differential. Thus, 
the differentials of logo?, sin -1 x, tan" 1 ^, etc., are algebraic func- 
tions. Moreover, it is evident that certain forms of differentials 
do not arise from the differentiation of any known functions. 
Hence, to obtain the exact integrals of many differentials, new 
and higher functions must be invented and studied. The 
integrals of these differentials, as obtained by series, are the 
developments of these, as yet, unknown functions. 



CHAPTER XVI. 

LENGTHS AND AREAS OF PLANE CURVES, AREAS OF SUR- 
FACES OF REVOLUTION, VOLUMES OF SOLIDS. 

195. Examples in Rectification of Plane Curves. For formulas, 
see § 65. 

1. Find the length of the parabola y 2 = 2px. 

Here s =/( i+ S) w= /( i+ D^ = ^ (y+2 ' 2)4d2/; 

.-. s = ^EEl + E\o g (y+ v9+7)+ C. § 185, Ex. 14. 

If s be measured from the origin, s = when y = 0, and 
G = — iplogp. 

2. Rectify the circle a?-{- y 2 = r 2 . 

We use L to represent the entire length of any closed curve. 

Here L = 4 f Yl + ^fdx = 4r f " dx = 2 rrr. 

Jo \ da? J Jo Vr 2 - x* 

For the value of 7r, see § 107. 

3. Rectify the ellipse y 2 = (1 - e 2 ) (a 2 - a 2 ) . 
Here ^ = - (1- e 2 ) * = _^IE? ; 

/» « <fcc / e 2 ^ eV 3e 6 a 6 \ 

= 4 JoV^=^V 2 « 2 ' 4 « 3 2.4.6a 5 "V 



RECTIFICATION OF PLANE CURVES. 



219 



r* a dx 2e 2 r a x 2 dx e 4 C a x'dx 

Jo Va 2 -r a J° Va 2 - x 2 2 a 3 Jo y a^ 

3e 6 /» tt x 6 dx 
2 -6a 5 Jo V^^? 



For the indefinite integrals of the last three terms, see § 185, 
Examples 1, 2, and 7. Finding the definite integrals between 
the given limits, and adding the results, we have 

3e 4 3 2 -5e 6 



2^1--- 



2 2 2 2 • 4 2 2 2 • 4 2 • 6 2 



4. Rectify the hypocycloid a# -f^ = at. 

5. Rectify the tractrix. 

The characteristic pro- 
perty of the tractrix is that 
the length of its tangent pt 
is constant. Denote this 
constant length by a ; and 
let o be the origin, oa being 
the tangent at a ; then, if 
pm= ds, —T^=dy, NM=C?£, 



Ans. 6 a. 




et = Va 2 — 2/ 2 . 



T -r ds PM 

Hence — = = 



dy 



,'.s 



also, -^ = 
dx 



Fig. 54. 



/ 9 

Vet"* — 2/" 



(1) 



— a I — = a 






if s be measured from a. 



In this example we have found the length of a curve without 
knowing its equation. For the length of the catenary obtained 
in a similar way, see § 76, Ex. 8. 

196. To rectify a curve given by its polar equation. 

Since ds 2 = p 2 dO 2 + dp 2 , § 136 (3) . 



220 LENGTHS AND AREAS OF PLANE CURVES. 

Examples. 

1. Rectify the spiral of Archimedes, p = a6. 

Here — = - ; .* . s = - I (cir 4- p 2 ) * dp. 
dp a aJ 

Hence, if s be measured from the pole, we have 

p(a 2 + p 2 )5 a n p-j- Va 2 + p 2 mok r w 
2 a 2 & a 

This is equal to the arc of the parabola, y 2 = 2 ax, intercepted 
between the vertex and the point whose ordinate equals p 
(§195, Ex. 1). 

2. Rectify the logarithmic spiral p = a e . 

Here s= I (l4-m 2 )Mp = (l-\-m 2 )*p, 

in which m is the modulus of the system of logarithms whose 
base is a, and s is measured from the pole. 

3. Construct and rectify the cardioid p = a(l+ cos0). 

L = 2 f[a 2 (l+ cos0) 2 + « 2 sin 2 0]sd0 

Jo 2 2 

since 2(1 + cos (9) = 4 cos 2 — 

z 

197. Examples in Quadrature of Plane Curves. The quadra- 
ture of a figure or surface is the finding of its area. For 
formulas, see §66. 

1 . Find the area of the circle x 2 + y 2 = r 2 . 
Here area = 4 J (r 2 — x 2 )idx 

nr =7r?- 2 . §185, Ex. 8. 



= 4p 



r 2 — x 2 )* , r 2 . _->x 

'—4 sm - 

2 2 7- 



EXAMPLES. 221 

For the segment between the lines x = a and x = 6, we have 
= 2 I (r 2 — x 2 ) h dx 



area 



= b Vr 2 — 6 2 + J* 2 sin -1 7 — aVr 2 — a 2 — r^sin -1 - 
2. Find the area of one branch of the cycloid. 



ii 

x = r vers -1 - — v2ry — y 2 . 

Area =2 C y ' dy = 3^. § 185, Ex. 18. 

Jo V2?'2/-2/ 2 

Hence the area of one branch is three times that of the gen- 
erating circle. 

3. Find the area of the tractrix. 



Here &-_, y - ,orcfce=-^- — ^cty. §195, Ex. 5. 

da; Va 2 - / 2/ 

.•.z=jydx = — j Va 2 — tfdyi 

.'. area = - 4 C\/a 2 -y 2 dy = ira 2 . § 185, Ex. 8. 

Hence the whole area enclosed by the curve is equal to the 
area of a circle whose radius is a. 

X s 

4. Find the whole area between the cissoid y 2 = — and 

., , , 2a — x 

its asymptote. 

Ans. 37T<r. 
# 

5. Fmd the area between the lines xry = a 3 , x = 5, x = c, and 

y = °- Ans. a* b -^> 

be 

6. Find the area of both loops of the curve tfy 2 = aP&aP—tiPx*. 

Ans. %ab. 

7. Find the area of one loop of the curve aV = x 4 (a 2 — x 2 ) . 

Ans. ia 2 . 



222 LENGTHS AND AREAS OF PLANE CURVES. 

198. To find the area of a curve given by its polar equation. 
Let p be any point on the curve ab referred to the pole o and 
b the polar axis ox. Take od=1, and draw 

the arcs db and pr ; then, if db = dO, it is evi- 
dent that the sector opr = cL4, A representing 
... ; h the area traced by the radius vector. 

p Hence dA = ^pr-op = ip 2 d0 ; 

a 




o x 



Fig. 55. .'. A — \ j p 2 d$. 

Or, let dc = AO ; then kp'= A/>, and area opp'= AA. 
Now opk < opp'< OHP f , or £-A0 < AA< ^ P + Ap ^ AO. 

2 AO 2 dO 2 ' V H 



Examples. 

1. Find the area of the first spire of the spiral of Archimedes, 
p = aO ; also the area between the first spire and the second. 



Here A = £ Ca 2 2 dO = £ a 2 3 = ±p 2 0. 



(7=0, since, if the area be estimated from the pole, A = 
when = 0. 

When0=27r, p = r, or the radius of the measuring circle; 
and A = ^irr 2 . Hence the area of the first spire is one-third 
of the area of the measuring circle. 

When = 4 ?r, p = 2r, and A = f -rvr 2 . 

But, in »the two revolutions, the area of the first spire has 
been traced twice ; hence the area between the first spire and 
the second is \irr 2 — \-irr 2 , or twice the area of the measuring 
circle. The area between the second spire and the third is four 
times the area of the measuring circle ; and so on. 

2. Find the area of the curve p = a sin 30. 
The curve consists of three equal loops (Fig. 47). Hence 
the area equals three times the area of the first loop. 

Ans. %Tra 2 . 



QUADRATURE OF SURFACES OF REVOLUTION. 223 

3. Find the area of the lemniscate p 2 = a 2 cos 2 6. 

The integral between = and = ^tt is one-fourth of the 

whole area. A 2 

Ans. or. 



4. Find the area of the cardioid p = a (cos 6 + 1 ) . 



Ans. ^ttcl 2 . 



199. Examples in Quadrature of Surfaces of Revolution. For 

formulas, see § 68. 

1 . Find the area of the surface of the prolate spheroid ; that 
is, of the surface traced by the revolution of the ellipse 

y' = (l-e 2 )(a 2 -x 2 ) 

about the axis of x. 

Area = 2 Clir^l- e 2 )l{a 2 -x 2 )\fl + ^fdx 

= 4 ve (i _ e 2) I f Y «j _ aA 4 d« 

= 47re- I f — — scM'cfa? 

= 4^[W^ - tf+fa^ZiY 1 185, Ex.8. 

tt6 2 H sin 2 e. 

e 

2. Find the area of the surface of the prolate spheroid whose 
generatrix is 9?/ 2 + 4a? 2 = 36. 

3. Find the area of the surface generated by the revolution 
of the cycloid about its base. 

Area = 2 £ 2 iryds = 4 * £y(l + ^*dy 

= 4ttV27' f y(2r-y)~idy 

= 4tt V27 [-f (4r+2/) (2r-y)*]! r § 179, Ex.3. 



= -§#-7T?' 2 . 



224 



CUBATURE OF SOLIDS OF REVOLUTION. 



4. Find the area of the surface generated by the catenary 
revolving about the axis of x, between the limits and b. 



Here S=27rj yds = ira\ [e a -\-e a ]ds 

= |7raJ o \e a + e~ a )dx §76, Ex. 8. 

= 7r [f(^- e "^)+^]. 

5. Find the area of the surface generated by the revolution 
of the tractrix about the axis of x. (See § 195, Ex. 5.) 

Ans. 47ra 2 . 



200. Examples in Cubature of Solids of Revolution. The 

cubature of a solid is the finding of its volume. For formulas 

see § 69. 

1. Find the volume of the solid generated by the revolution 
of the cycloid about its base. 

ydy . o 7 7ry 3 dy 

V2 ry — y 2 V2 ry — y 2 



Here 



dx 



.*. volume 



= 2tt f- 



y s dy 



= 5' 



§185, Ex. 19. 



■V2ry-y 2 
that is, the volume is five-eighths of the circumscribed cylinder. 

2. Find the inclosed volume of the solid generated by the 
revolution of the parabola y 2 = 2px about the line x = a. 

LetMKbe the line x=a ; let&=AK[= -y/2pa] ; 
and let p be any point on the parabola ; then 
oh — x, and ha = a— x. Now, if bc = dy, the 
volume generated by the revolution of bpdc 
about mk equals dV\ 

.'. dV= ?r(a — x) 2 dy ; 

.-. volume = 2?r | ( a— $— ) dy = j-f -rrba 2 . 




Fig. 56. 



3. Find the volume of the solid generated by the revolution 
of the cissoid about its asymptote. Ans. 2tt 2 o?. 



EXAMPLES. 225 

4. Find the volume of the solid generated by the revolution 
of the tractrix about the axis of x. 

Since y = a when x = 0, and y = when x = oo, and 



■Va 2 -f 

y 

volume = 2 I Try 2 dx = — 2 ir j ?/ V a 2 — y 2 dy = % -n-a 3 . 

c/x=0 «/y=a 



eta^- V" -y d y; 

y 

»y=0 



201. The Calculus is often of great aid in deducing the equa- 
tions of curves. The equation of the catenary is obtained by 
its use in § 76, Ex.8. 

Examples. 

1. Find the equation of the tractrix. 

Here ^ = - 9 V ', § 195, Ex. 5. 

ax \w — y~)* 

J y 

.-. x=a log a + (ft2 ~ y2) * - (a 2 - y 2 )h. §185, Ex. 13. 
G— 0, since a? =0 when y= a. 

2. Find the equation of the curve whose subtangent is c. 

Here y — [=subt.] = c ; 
dy 

.-. dx = c~; 

y 

.-. z = -log a 2/+C. 
m 

If c — m^ and the curve pass through the point (0, 1), we 

have , 

x = log a y. 

3. Find the equation of the curve whose subnormal is c times 
the square of its abscissa. A 2 ? 3 1 n 



CHAPTER XVII. 
THE METHOD OF INFINITESIMALS. 

202. Infinitesimals and Infinites. A quantity so small that its 
value cannot be expressed io terms of a finite unit, is said to be 
infinitely small. 

An Infinitesimal is an infinitely small variable whose limit is 
zero. For example, if y =f(x) , Ax and Ay both become infini- 
tesimals as Ax = 0. Again, any variable, when near its limit, 
differs from its limit by an infinitesimal. 

When we consider several related infinitesimals, we choose 
arbitrarily some one of them as the principal infinitesimal, and 
adopt the following definitions : 

Any infinitesimal, the limit of whose ratio to the principal 
infinitesimal is finite, is an infinitesimal of the first order. 

Any infinitesimal, the limit of whose ratio to the square of the 
principal infinitesimal is finite, is an infinitesimal of the second 
order. 

Any infinitesimal, the limit of whose ratio to the nth power 
of the principal infinitesimal is finite, is an infinitesimal of the 
nth order. 

Hence, if t represent the principal infinitesimal, v x l, v 2 t 2 , v 3 l 3 , 
and v n i n will represent respectively any infinitesimals of the first, 
second, third, and nth orders, in which v x , v 2 , v 3 , and v n are 
variables having finite limits, from which they differ by infini- 
tesimals. According to this notation, a = i, a = v 1 L, and a = v 2 t 2 
are read respectively, " a = the principal infinitesimal," " a = an 
infinitesimal of the first order," and "a = an infinitesimal of the 
second order." 

A quantity so large that its value cannot be expressed in 
terms of a finite unit, is said to be infinitely large. 



GEOMETRIC ILLUSTRATION. 227 

An Infinite is an infinitely large variable that increases with- 
out limit. Hence the reciprocals of infinitesimals are infinites, 
and the different orders of infinites may be represented by iv 1 r l , 
w 2 r 2 , w s r 3 y etc., in which w 1? w 2 , w 3 , etc., are variables having 
finite limits. 

The symbols and oc, or -, represent respectively absolute 

zero and absolute infinity, of which there are no orders. 

203. From the algebraic symbols for infinitesimals and infi- 
nites of different orders, the following principles are evident : 

1. The product of any infinitesimal and an infinite of the 
same order is a finite quantity ; thus, v 2 t 2 • iv 2 r 2 = v 2 w 2 . 

2. The order of the product of two or more infinitesimals is 
the sum of the orders of the factors ; thus, v x l'V 2 l 2 = v 1 v 2 l 3 . 

3. The order of the quotient of any two infinitesimals is the 
order of the dividend minus the order of the divisor ; thus, 

V 2 L 2 v 2 

4. If the limit of the ratio of one infinitesimal to another is 
zero, the former is of a higher order than the latter ; thus, 

limit ^i 2 = limit-^- t = 0. • 

ViL v ± 

204. Geometric Illustration of Infinitesimals of Different 
Orders. Let cab be a right angle inscribed in the semicircle 
cab, bd a tangent at b, and ae a perpendicular to bd. From 
the similar triangles cab, bad, and aed, ^d 
we have 

AD AB , 1 s 

= >, (I) 

AB AC 

and E5 = -^. (2) 

AD BC 




Fig. 5 



Suppose a to approach b so that ab = t. Since limit ab = 0, 
and limit ac = cb, from (1) we have 



228 THE METHOD OF INFINITESIMALS, 

limit — = limit — = ; 

AB AC 

hence ad is an infinitesimal of a higher order than ab (§ 203, 4). 
From (2) we have 

limit — = limit — = ; 

AD BC 

hence de is an infinitesimal of a higher order than ad. 
Thus, when ab = t, ad = v 2 t 2 , and de = v 3 t 3 . 

Examples. 

1 . If a = i, of what order is sin a ? 

Since H T* ["*Eif] =; 1 , & in « = Vi L .'(§ 202) , 

and limit v 1 = 1 . 

2. If a = t, of what order is tan a? 

3. If a = t, of what order is 1 — cos a? 

Since ^ P ~ C 2 0S a l = f , 1 - cos a = V (§ 202) , 

u • - 1 

and limit v 2 = -J. 

4. If a = t, show that sin a— a=v 3 t 3 , and that a— tana= / y 3 t 3 . 

205. First Fundamental Principle of Infinitesimals. 

Let a — fi = c, in which e is infinitely small in comparison with 
a or /5 ; then 

2L-— l+.i, andlimit| = 0; 
P P " © 

\ limit - = limitfl + l~j = 1. 

Hence, if the difference between two variables is infinitely small 
in comparison with either of them, the limit of their ratio is unity; 



RULE FOR DIFFERENTIATION. 229 

and, by § 135, either of them may be substituted for the other in 
any problem concerning the limit of the ratio of tioo variables. 

For convenience of application, this principle may be stated 
as follows : 

In problems concerning the limit of the ratio of two variables, 

All infinitesimals of the higher orders may be dropped from 
sums of infinitesimals of different orders. 

All infinitesimals may be dropped from sums of finite quantities 
and infinitesimals. 

All finite quantities maybe dropped from sums of infinites and 
finite quantities. 

r limit 1 

when € is infinitely small in comparison with (3. 

Hence, conversely, if the limit of the ratio of tivo infinitesimals 
is unity, their difference is infinitely smcdl in comparison ivith 
either. 

206. Rule for Differentiation. In this chapter we shall regard 
the increments of variables as infinitesimals. Since the differ- 
ence between a variable and its limit is an infinitesimal, and 
since 



Cor. If a = /3 -f-e, limit - , or limit 1 + - 



is unity only 



a[S]-w 



.•.^=/'(a0 + e , ovAy=f'(x)Ax + e±x, (1) 

in which e is an infinitesimal. 

Now dy = f'(x)dx. (2) 

The value of dx being arbitrary, for convenience we shall, in 
this chapter, suppose it to be equal to Aa;. 

From (1) and (2) it follows that, if dx = Ax = c, dy and A?/ 
are infinitesimals whose difference is infinitely small in compari- 
son with either, and therefore, in differentiating, dy may be 
substituted for Ay (§ 205) . 



230 THE METHOD OF INFINITESIMALS. 

From these considerations we have the following simple rule 
for differentiating any function : 

Find the increment of the function in terms of the increments 
of its variables, apply the principles of § 205, and in the terms 
remaining replace the increments by differentials. 

Thus, to differentiate X s , let y = x 3 ; then 

Ay = 3 x 2 Ax + 3 x{Ax) 2 -f (Aaj) 3 . 
Hence, by the rule, 

dy = 3 x 2 dx. 

Rem. We do not drop the infinitesimals of the higher orders, 
because the} T are nothing, or comparatively nothing, when added 
to an infinitesimal of the first order, but because we know that 
they do not appear in the limit of the ratio sought. Thus the 
method of limits is the basis of the method of infinitesimals, the 
difference being that in the latter we use infinitesimal differen- 
tials, and a quantity is dropped as soon as it appears, when it is 
known that it will vanish in passing to the limit sought. Any 
differential equation obtained by the infinitesimal method must 
evidently be true when the differentials are regarded as finite. 

Examples. 

1 . Differentiate u = xy. 

Here Au = y Ax + x Ay -f- Ax Ay ; .*. du = ydx -f- xdy. 

x 

2. Differentiate u = — 

y 

Here Au = V**~**y • .-.du = V^-^V . 
y 2 +yAy y 2 

3. Differentiate y = sin.T. 

Here Ay — sin (x -j- Ace) — sin x 

= cos x sin Ax — { 1 — cos Ax) sin x 
== cosxAx-t~v 2 L 2 ; 



SECOND FUNDAMENTAL PRINCIPLE. 231 

for, when Ax = i, Ax = sin Ax + v 2 r(§ 48, Cor., and § 205, Cor.), 
and l-cosA£C = <y 2 i 2 (§ 204, Ex. 3). 

. • . dy = cos x dx. 

4. Find the differential of any plane curve. ^J, 

Let As, or arc pp' in Fig. 58, be t ; then 

As = chord pp'-f- v 2 i 2 § 48. 

= ^/Ax 2 + Ay 2 -}-v 2 l 2 ; 




.-.ds = V dy 2 + dx 2 . 

5. Find the differential of the area between a curve and the 
axis of x. 

Let Ax, or ab in Fig. 58, be i ; then, since area pdp'< Ax* Ay, 
Az = area abp'p = y Ax + v 2 i 2 ; 
,'.dz = ydx. 

207. Second Fundamental Principle of Infinitesimals. Let 

a i5 a 25 a 3? •*•? a n t> e anv infinitesimals so related that, as n 
increases, 

limit [«! + a 2 + a 3 H |- a n ] = c ; 



and let ft, f3 2 , (3 3 , ••» /?„ be any other infinitesimals, such that 

^ = l+ q , £- 2 =l-bc 2 , ..-, ^=l + €n , (1) 

ttl a 2 a n 

in which e t , e 2 , •••, c n are infinitesimals. 

Clearing equations (1) of fractions, adding, etc., we obtain 

/?! + &+••• +/?n-(a 1 + a 2 +...+a n ) 

= a 1 € 1 -\-a 2 e 2 -\ f- a»e». 

Let 8, a positive infinitesimal, be greater in absolute value 
than any of the infinitesimals e 1? e 2 , •••, e n ; then we have numeri- 
cally, 

< 8 ( ai + a 2 H f- a n ) . 



232 THE METHOD OF INFINITESIMALS. 

But, since limit 8 = 0, and limit . ( a i + «2 H h <0 = c, 

limit [8( ai + a 2 H 1- «„)] = 0. 

Whence limit [ft + ft -\ h ft] = limit [a x + a 2 H h a n ] . 

Hence, if the difference between two infinitesimals is an infini- 
tesimal of a higher order, either may be substituted for the other 
in any problem concerning the limit of the sum of infinitesimals, 
provided this limit is finite. 

Cor. If - = !+€!, — =l + e 2 , •••, — = 1 + e«, 

a l a 2 a n 

and a x + a 2 -\ \-a n = C, 

then limit [ft 4- ft H h ft] = «i + « 2 H h «« = c. 

208. Integration as a Summation. Let 2 represent the area 
between the curve opd and the axis of x, and let us seek the 
area of the portion obd. At the several 
values of x, as, 0, oa, oa', and oa", let 
oa = aa' = a'a" = a' f B = Ace = dx ; 
then the corresponding values of dz are 
0, apba', a'p'b'a", and a"p"6"B, while those 
x of Az are opa, appV, a'p'p"a", and a"p"DB, 
rig. 59. whose sum equals obd. Let the divisions 

oct, aa', etc., become infinitesimals, bat increase in number so 
that their sum will pontinually equal ob ; then Az and dz both 
become infinitesimals; and, since op«, pp'5, etc., is each less 
than Ax • Ay, Az — dz = v 2 r. Therefore the limit of the sum 
of the values of dz is equal to the sum of the values of Az 

x = x' 

(§ 207, Cor.). Hence, if x' =ob, and the symbol ]£ Az repre- 
ss 

sent the sum of the values of Az corresponding to the different 
values of x between and x', we have 




Xz=X 



area obd = ^ Az = limit % dz = limit £ ydx. 

x=0 x=0 

J ydx; 



x=0 
f*x' 

But area obd = 



•1 



ydx = limit J yefce. 



CENTRE OF GRAVITY. 233 

Hence, ivhen differentials are infinitesimals, integration may 
be viewed as the summation of an infinite series of infinitesimals. 

209. Centre of Gravity. The centre of gravity of a body is a 
point so situated that, if it be supported, the bocly will remain at 
rest in whatever position it may be placed. An element of any 
quantity or magnitude is an infinitely small portion of it. The 
product of the weight of a body by the distance of its centre of 
gravit}' from a given plane is called the moment of the bod} T with 
respect to that plane. The moment of a bod}' is the sum of the 
moments of its elements. Hence the distance of the centre of 
gravity of a body from a given plane equals the sum of the 
moments of its elements divided \>y the weight of the body. 
The bodies here considered are supposed to be of uniform den- 
sity ; hence their weights are proportional to their volumes. 

The advantage sometimes gained b} 7 viewing integration as a 
summation is illustrated in deducing formulas for finding the 
centre of gravity. 

210. To find the centre of gravity of any plane surface. Let 
(x, y) be any point on the curve ovn referred to the axes ox 
and oy, and let x and y represent respec- 
tively the distances of the centre of gravity of 
any portion of the surface xon from the planes 
oy and ox, which planes are perpendicular to 
that of the figure. The differential of the 
area xo?i is ydx ; now. if dx = ab = i, ydx will 
differ from abp'p, the corresponding element Flg ' 60 ' 

of this area, by v 2 c : and the distance of the centre of gravity 
of this element from the plane oy will differ from x hyv 1 c; 
hence xydx will differ from the moment of this element with 
respect to the plane oy, by v 2 r. Therefore, between x = a and 
x = b, the sum of the moments of the elements 

= limit % xy dx = j xy dx ; 

x=a *Ja 



area 



Xb r*b 

xydx j xydx 

J ydx 



234 THE METHOD OF INFINITESIMALS. 

Again, the centre of gravity of ydx is evidently \y from the 
plane ox ; hence iy 2 (lx will differ from the moment of the cor- 
responding element with respect to the plane ox, by v 2 l 2 . Hence, 
between x = a and x = b, the sum of the moments of the ele- 
ments = \ I y 2 dx ; 

Xb fb 

y 2 dx I y 2 dx 



area 



Jydi 



If the curve be symmetrical with respect to ox, x is evidently 
the same for the whole area as for the half, and y is zero. 

211. To find the centre of gravity of any plane curve. Let 
(x 01 y ) be the centre of gravity of any arc of a plane curve 
whose length is represented by s. Now, when ds = t, xds dif- 
fers from the moment of the corresponding element of the curve, 
with respect to the plane oy (Fig. 60) , by v 2 t 2 . Hence, between 
x = a and x = b, the sum of the moments of the elements 



= J xds, 
and \%ds 



Xn — 



In like manner, we obtain 



2/o = 



S) d ' 



212. To find the centre of gravity of a solid of revolution. 
The differential of a solid of revolution whose axis is the axis 
of x, is ny 2 dx ; hence, if dx = t, irxy 2 dx will differ from the 
moment of the corresponding element of the solid, with respect 
to the plane ot, by v 2 i 2 ; and therefore, between x = a and x = b, 
the sum of the moments of the elements 

= I 7rxy 2 dx. 



EXAMPLES. 



235 



7T I xy 2 dx I xy 2 dx 
I yhlx 

4/a 



Xq — ■ 



volume 



As the centre of gravity must evidently be on the axis of revo- 
lution, the formula given above entirely determines it. 



Examples. 

1. Determine the centre of gravity of a circular arc bad. 
Let the extremity d be (x' r y'). 
Here y 2 = 2 rx — x 2 ; 
(r—x)dx 



dy 



•\j2rx — x 2 
= Vcfce 2 + dy 2 = 



rdx 



V2 



rcc — xf 



,\ ae = a? 



jf 




C?8 



Fig. 61. 









sJ°V2r£ — a 2 * 



= I[_V2ra'-^' 2 + s] 






ry 



Hence ce = r — ae 



r y _ r chord bd 



arc bap 



2. Find the centre of gravity of a segment of a circle. 
Using the equation of the circle referred to its centre, we have 

xydx j (r 2 — x 2 )hxdx 
area area 



area 



If a = 0, and b = r, then area == Jtt?* 2 , and we have # = — - 
when the segment is a semicircle. 



236 THE METHOD OF INFINITESIMALS. 

3. Find the centre of gravity of a parabolic area. 

Ans. x Q = ^x'. 

4. Find the centre of gravity of a right cone. 
Here y = ax, and volume = ^iry 2 x ; 

7r I xy 2 dx tt J a 2 x s dx 

volume ■§Tra z x 16 

that is, the distance of the centre of gravity from the vertex is 
three-fourths of the axis. 

5. Find the centre of gravity of a segment of a prolate 

spheroid. ^^'-i>) 

Ans. x — - 

volume 

When x'= a, x = J- a. 

I xyds 

6. Prove that # =^ is tne formula for finding the cen- 

iyds 

tre of gravity of any surface of revolution. 



